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Amount at the end of the 1st year = Rs 1100 

Amount at the end of the 2nd year = Rs 1210 

Amount at the end of 3rd year = Rs 1331 and so on. 

So, the amount (in Rs) at the end of 1st year, 2nd year, 3rd year, ... are 

1100, 1210, 1331, ... 

Here, a₂ – a₁ = 110 

a₃ – a₂ = 121 

As, a₂ – a₁ ≠ a₃ – a₂ , it does not form an AP.

Let the required sum = a

and the interest carried every year = d

According to question,

In 4years, a sum of money kept in bank account = Rs. 600

i.e. t₅ = 600 ⇒ a + 4d = 600 …(i)

and in 12 years , sum of money kept = Rs. 800

i.e. t₁₃ = 800 ⇒ a + 12d = 800 …(ii)

Solving linear equations (i) and (ii), we get

a + 4d – a – 12d = 600 – 800

⇒ – 8d = –200

⇒ d = 25

Putting the value of d in eq.(i), we get

a + 4(25) = 600

⇒ a + 100 = 600

⇒ a = 500

Hence, the sum and interest carried every year is Rs 500 and Rs 25 respectively.

Amount of money = 700 

Total number of prize = 7 

Each prize is Rs. 20 less than its preceding prize Let, 

Prize of first prize = a 

Prize of second prize = a – 20 

Prize of third prize = a – 20 – 20 = a – 40 

Here, A.P. is a, a – 20, a – 40,… 

So, common difference, d = a – 20 – a = -20

We Know, S_n = \(\frac{n}{2}\)[2a + (n-1)d]

700 = \(\frac{7}{2}\)[2a + (7-1)-20]

\(\frac{1400}{7}\) =2a + 6(-20)

2a  = 200 + 120

2a = 320

a = 160

so, value of first prize = 160

Value of second prize, 160 - 20 = 140

Value of third prize, 140 - 20 = 120

Value of fourth prize 120 - 20 = 100

Value of fifth prize, 100 - 20 = 80

Value of sixth prize, 80 – 20 = 60 

And value of seventh prize, 60 – 20 = 40 

Hence, value of prizes, Rs 160, Rs 140, Rs 120, Rs 100, Rs 80, Rs 60, Rs 40.

x-y/x+y,3x-2y/x+y,5x-3y/x+y

For the given AP the first term a is (x - y)² and common difference d is a difference of the second term and first term, 

Which is \(\frac{3x-2y}{x+y}\) - \(\frac{x-y}{x+y}\) = \(\frac{2x-y}{x+y}\)

To find : the sum of given AP 

The formula for sum of AP is given by,

s = \(\frac{n}{2}\)(2a+(n-1)d)

Substituting the values in the above formula,

s = \(\frac{n}{2}\)(\(2\frac{x-y}{x+y}\)+(n-1)\((\frac{2x-y}{x-y})\))

Given terms of A.P = (x – y), (x + y), (x + 3y) 

then common difference = difference of successive terms 

= (x + y) – (x – y) = x + y – x + y = 2y 

∴ Common difference of given AP = 2y

In the given A.P., a₁ = 117, a₂ = 104 

Common difference d = a₂ – a₁ 

= 104 – 117 

= -13 

∴ 8^th term t₈ = a₁ + 7d

= 117 + 7(-13) 

= 117 – 91 = 26

The terms are a, ar, ar², ar³, …….. 

∴ 4, 4 × 3, 4 × 3², 4 × 3², …… 

⇒ 4, 12, 36, 108, ……

(i) a = 7, d = 3, n = 8, a_n =? 

a_n = a + (n – 1)d 

a₈= 7 + (8 – 1)3 

= 7 + 7 × 3 = 7 + 21 

∴ a₈ = 28 

(ii) a = -18, d =?, n = 10, a_n = 0 

a_n = a + (n – 1)d 

0 = -18 + (10 – 1)d 

0 = -18 + 9d 18 = 9d 

9d = 18

∴ d = \(\frac{18}{9}\) = 2

(iii) a =?, d = -3, n = 18, a_n = -5 

a_n = a + (n – 1)d 

-5 = a + (18 – 1) (-3) 

= a + 17(-3) 

-6 = 1 – 51 

∴ a = -5 + 51 = 46 

(iv) a = -18.9, d = 2.5, n =? a_n = 3.6 

a_n = a + (n – 1) d 

3.6 = -18.9 + (n – 1) (2.5) 

3.6 = -18.9 + 2.5n – 2.5 

3.6 = 2.5n – 21.4 

2.5n = 3.6 + 21.4 

2.5n = 25 

n = \(\frac{25}{2.5}\) = \(\frac{250}{25}\)

∴ n = 10. 

(v) a = 3.5, d = 0, n = 105, a_n =? 

a_n = a + (n – 1) d 

= 3.5 + (105 – 1) (0) 

= 3.5+ 104 × 0 

= 3.5 +0 

∴ a_n = 3.5

We know that the sum of terms for an A.P is given by

S_n = \(\frac{n}{2}\)[2a + (n − 1)d]

Where; a = first term for the given A.P. d = common difference of the given A.P. n = number of terms

Or S_n = \(\frac{n}{2}\)[a + l]

Where; a = first term for the given A.P. ;l = last term for the given A.P

Given series, 25 + 28 + 31 + . . . + 100 which is an A.P

Where, a = 25 ,d = 28 – 25 = 3 and last term (a_n = l) = 100

We know that, a_n = a + (n – 1)d

So,

100 = 25 + (n – 1)(3)

100 = 25 + 3n – 3

n = \(\frac{(100 \,–\, 22)}{3}\) = 26

Now, for the sum of these 26 terms

S₁₀₀ = \(\frac{26}{2}\) [24 + 100]

= 13(124)

= 1625

Hence, the sum of terms of the given series is 1625.

Given an AP whose kth term is 5k + 1 

To find : the sum of n terms of an AP whose kth term is 5k + 1 

So substituting k = 1 to get the first term a₁ = 6 

Substituting k = 2 to get the second term a₂ = 11 

d = a₂ - a₁ = 5 

We know that the sum of AP is given by the formula :

s = \(\frac{n}{2}\)(2a + (n-1)d)

Substituting the values in the above equation,

s = \(\frac{n}{2}\)(12 + (n-1)5)

s = \(\frac{5n^2+7n}{2}\)

Given the sum of the certain number of terms of an A.P. = 116

We know that, S_n = \(\frac{n}{2}\)[2a + (n − 1)d]

Where; a = first term for the given A.P.

d = common difference of the given A.P.

n = number of terms So for the given A.P.(25, 22, 19,…)

Here we have, the first term (a) = 25

The sum of n terms S_n = 116

Common difference of the A.P. (d) = a₂ – a₁ = 22 – 25 = -3

Now, substituting values in S_n

⟹ 116 = \(\frac{n}{2}\)[2(25) + (n − 1)(−3)]

⟹ (\(\frac{n}{2}\))[50 + (−3n + 3)]  = 116

⟹ (\(\frac{n}{2}\))[53 − 3n] = 116

⟹ 53n – 3n² = 116 x 2

Thus, we get the following quadratic equation,

3n² – 53n + 232 = 0

By factorization method of solving, we have

⟹ 3n² – 24n – 29n + 232 = 0

⟹ 3n( n – 8 ) – 29 ( n – 8 ) = 0

⟹ (3n – 29)( n – 8 ) = 0

So, 3n – 29 = 0

⟹ n = 29/3

Also, n – 8 = 0

⟹ n = 8

Since, n cannot be a fraction, so the number of terms is taken as 8.

So, the term is:

a₈ = a₁ + 7d = 25 + 7(-3) = 25 – 21 = 4

Hence, the last term of the given A.P. such that the sum of the terms is 116 is 4.

Answer is (A) 89

a₇ = 34 (given)

a₁₃ = 64 (given)

– 6d = -30

d = 5

put the value of d in a + 6d = 34

a + 6 × 5 = 34

a = 34 – 30

a = 4

∴ a₁₈ = a + 17d

= 4 + 17 × 5 = 4 + 85 = 89

Given, 

First term, a = 22

n^th term, a_n= -11 = l

and sum of n terms, S_n= 66 

We know sum of n terms, S_n = \(\frac{n}{2}[a + l]\)

66 = \(\frac{n}{2}[22 - 11]\)

66 x 2 = n x 11

n = \(\frac{66 \times 2}{111} = 12\)

we know, a_n= a + (n – 1) d -11 

= 22 + (12 – 1) d -11 – 22 

= 11d -33 = 11d 

d = -3

Hence, number of terms, n = 12 and common difference, d = -3

Let n be the total number of terms and d be the common difference.

Given:

first term = a = 22

nth term = -11

Sum of all terms = S_n = 66

S_n = n/2[a + l]

66 = n/2[22 + (-11)]

n = 12

There are 12 terms in the AP.

Since nth term is -11, so

a_n = a + (n – 1)d

-11 = 22 + (12-1)d

d = -3

Therefore, Common difference is -3 and the number of terms are 12.

Given: a_n = 5 – 6n

Find some of the terms of AP:

Put n = 1, we get a₁ = – 1 = first term

Put n = 2, we get a₂ = – 7 = second term

Common difference = d = a₂ – a₁ = – 7 – (-1) = – 6

Sum of first n terms:

S_n = n/2 [2a + (n – 1)d]

= n/2[- 2 + (n – 1)(-6)]

= n(2 – 3n)

sum of first 20 terms:

S₂₀ = 20/2[2(-1) + (20 – 1)(-6)]

= 10 [ – 2 – 114]

= – 1160

Sum of its first 20 terms of AP is -1160.

Let’s take the first term as a and the common difference to be d

Given that,

a₅ = 30  and a₁₂ = 65

And, we know that a_n = a + (n – 1)d

So,

a₅ = a + (5 – 1)d

30 = a + 4d

a = 30 – 4d   …. (i)

Similarly, a₁₂ = a + (12 – 1) d

65 = a + 11d

a = 65 – 11d …. (ii)

Subtracting (i) from (ii), we have

a – a = (65 – 11d) – (30 – 4d)

0 = 65 – 11d – 30 + 4d

0 = 35 – 7d

7d = 35

d = 5

Putting d in (i), we get

a = 30 – 4(5)

a = 30 – 20

a = 10

Thus for the A.P; d = 5 and a = 10

Next, to find the sum of first 20 terms of this A.P., we use the following formula for the sum of n terms of an A.P.,

S_n =\(\frac{ n}{2}\)[2a + (n − 1)d]

Where;

a = first term of the given A.P.

d = common difference of the given A.P.

n = number of terms

Here n = 20, so we have

S₂₀ = \(\frac{20}{2}\)[2(10) + (20 − 1)(5)]

= (10)[20 + (19)(5)]

= (10)[20 + 95]

= (10)[115]

= 1150

Hence, the sum of first 20 terms for the given A.P. is 1150.

Correct anwser is (d) -b

The given AP is \(\frac{1}{3},\frac{1-3b}{3},\frac{1-6b}{3},.....\)

∴ Common difference, d = \(\frac{1-3b}{3}-\frac{1}{3}=\frac{1-3b-1}{3}=\frac{-3b}{3}=-b\)

All odd numbers between 0 and 50 are 1, 3, 5, 7,……… 49.

This is an AP in which a = 1, d = (3 - 1) = 2 and \(l=49.\)

Let the number of terms be n.

Then, T_n = 49

\(\Rightarrow\) a + (n - 1) d = 49

 \(\Rightarrow\) 1 + (n - 1) x 2 = 49

\(\Rightarrow\) 2n = 50

\(\Rightarrow\) n = 25

∴ Required sum = \(\frac{n}{2}(a+l)\)

= \(\frac{25}{2}[1+49]=25\times25=625\)

Hence, the required sum is 625.

a = 50, d = 20 

a_n = a + (n – 1) d = l 

a₁₂ = 50 + 11 x 20 = 270 

S_n = \(\frac{n}{2}\)[a + l] 

= \(\frac{12}{2}\)[50 + 270] 

= 12 x 160 = 1920 

Yes, she will be able to fulfill her dream of sending her daughter to school.

The money saved by Sudha in the first year = Rs. 100

According to the question,

Sudha increased the amount by Rs. 50 every year

⇒ The money saved by Sudha in a 2^nd year = Rs. 100 +50

= Rs. 150

The money saved by Sudha in a 3^rd year = Rs. 150 + 50

= Rs. 200

Therefore, the series is

100, 150, 200, 250,…

Difference in the 2^nd term and 1^st term = 150 – 100 = 50

Difference in the 3^rd term and 2^nd term = 200 – 150 = 50

Since the difference is the same.

Therefore, the money saved by Sudha in successive years form an AP with a = Rs 100 and d =Rs 50

No, because the total fare after each km is

15, 15 + 8, 15 + 8(2), 15 + 8(3) …….

15, 23, 31, 39….

a₁ = 15, a₂ = 23, a₃ = 31 and a₄ = 39

a₂ - a₁ = 23 - 15 = 8

a₃ - a₂ = 31 - 23 = 8

a₄ - a₃ = 39 - 31 = 8

Since, all the successive terms of the given list have same difference i.e., common difference = 8.

Hence, the total fare after each km forms an AP.

The taxi fares are in A.P. with the first term 14 and common difference 2.

\(\therefore\) a = 14, d = 2, n = 10

Now, t_n = a + (n – 1) d

\(\therefore\) t₁₀ = 14 + (10 – 1) x 2

= 14 + 9 x 2

= 32

\(\therefore\) The taxi fare for 10 kilometres will be Rs.32.

Taxi fare for 1km = 15

According to question, Rs. 8 for each additional km

⇒ Taxi fare for 2km = 15 + 8 =23

and Taxi fare for 3km = 23 + 8 =31

Therefore, series is

15, 23, 31 ,…

Difference between 2^nd and 1^st term = 23 – 15 = 8

Difference between 3^rd and 2^nd term = 31 – 23 = 8

Since, difference is same.

Hence, the taxi fare after each km form an AP with the first term, a = Rs. 15 and common difference, d = Rs. 8

Given: -

Total debt = Rs.36000

A man pays this debt in 40 annual instalments that forms an A.P.

After annual instalments, that man dies leaving one - third of the debt unpaid.

So,

Within 30 instalments he pays two - thirds of his debt.

Sum of n terms in an Arithmetic Progression = n/2[2 x a+(n-1) x d]

He has to pay 36000 in 40 annual instalments,

36000 = 40/2[2 x a+(40 -1) x d] → (1)

Where,

a = amount paid in the first instalment,

d = difference between two Consecutive instalments.

He paid two – a third of the debt in 30 instalments,

2/3(36000) = 30/2[2 x a+(30 -1) x d] → (2)

From equations (1) & (2) we get,

a = 510 & d = 20

∴The value of the first instalment is Rs.510.

(i) 9 and 19

To find: Arithmetic mean between 9 and 19

The formula used: Arithmetic mean between a and b = +/2

We have 9 and 19

A.M. = 9+19/2

= 28/2

= 14

(ii) 15 and -7

To find: Arithmetic mean between 15 and -7

The formula used: Arithmetic mean between a and b = +/2

We have 15 and -7

A.M. = (-15)+(-7)/2

= 15-7/2

= 8/2

= 4

(iii) -16 and -8

To find: Arithmetic mean between -16 and -8

The formula used: Arithmetic mean between = a and b = +/2

We have -16 and -8

A.M. = (-16)+(-8)/2

= -16-8/2

= -24/2

= -12

ap = a => A + (p - 1) D = a ......(1)

aq = b => A + (q - 1) D = b .....(2)

ar = c => A + (r + 1) D = c ......(3)

Now, L.H.S. = a (q - r) + b(r - p) +c (p - q)

= {A + (p - 1)D} (q - r) + {A + (q - 1)D} (r - p) + {A + (r - 1)D} (p - q)

= 0. R.H.S

To Find: The sum of the given series.

Sum of the series is given by

S = n/2(a+l)

Where n is the number of terms, a is the first term and l is the last term

Here a = 101, l = 43, n = 30

S = 30/2[101+43]

= 15 × 144 = 2160

The sum of the series is 2160.

Total amount for counting = Rs. 10710 

In 1 min he counts =Rs. 180

In half an hour (30 min) he count = Rs.180×30 =Rs.5400 

Amount left count after half an hour, S_n= Rs.10710 – Rs. 5400 = 5310 

Now, in 31^st min he count 3 less than preceding minute = Rs.180 – Rs.3 = Rs.177 

In 32^nd min he count 3 less than preceding minute = Rs. 177 –Rs. 3 =Rs. 174 

Then, Arithmetic progression formed is 177, 174,… 

Here, difference between minutes = 174 – 177 = -3 

We know, sum of n terms, S_n= (n/2) [2a + (n – 1) d]

5310 = (n/2) [2(177) + (n – 1) -3] 

5310 × 2 = n [354 – 3n + 3] 

10620 = 354n – 3n² + 3n 

10620 = 357n – 3n² 

⇒ 3n² – 357n + 10620 = 0 

On taking 3 common from the complete equation, we get, 

⇒ 3 (n² – 119n + 3540) = 0 

⇒ n² – 119n + 3540 = 0 

Now, from factoring by splitting the middle term method, we get,

⇒ n² – 60n – 59n + 3540 = 0 

⇒ n (n – 60) – 59 (n -60) = 0 

⇒ (n – 59) (n – 60) = 0 

Therefore, either n = 59 or n = 60

We will use 59 as these are minutes So The total time to calculate whole amount is 59 + 30 = 89 min 

= 1 hr 29 min

Total numbers of trees = 25 

Distance between trees = 5 

Total distance between well and first tree is = 10 m

Now, gardener return back to well after watering each tree then, 

Distance between well and second tree is = 10 + 10 + 5 = 25 

Distance between well and third tree = 25 + 5 + 5 = 35 

Distance between well and fourth tree = 35 + 5 + 5 = 45 

Distance between well and last tree without return back = 10 + 24(5) = 10 + 120 = 130 

So, common difference = 35 – 25 = 10 

Now, Total distance covered by 24 trees,

We knew, S_n= \(\frac{n}{2}\)[2a + (n – 1) d]

= \(\frac{24}{2}\)[2(25) + (24 – 1) 10] 

= 12 [50 + 23 x 10] 

= 12 [50 + 230] 

= 12 x 280 = 3360

Now, distance covered by 25 trees = 10 + 3360 = 3370 

Total distance covered by gardener with return back = 3370 + 130 = 3500 m 

Hence, total distance will covered by gardener is 3500.

Hint:

Distances between trees and well are in A.P.

Given:

The distance of well from its nearest tree is 10 metres

Distance between each tree is 5 metres.

So,

In A.P

The first term is 10 metres and the common difference is 5 metres.

a = 10 & d = 5

The distances are in the following order

10, 15, 20… (30 terms)

The farthest tree is at a distance of a + (30 - 1) ×d

l = 10 + (29) ×5

L = 155 metres.

Total distance travelled by the Gardner = 2×Sum of all the distances of 30 trees from the well.

Sum of distances of all the 30 trees is n/2{a+1}

Sum = 30/2{10+155} metres

= 15 × 165 metres

= 2475 metres.

Total distance travelled by the Gardner is 2 × 2475 metres.

∴The total distance travelled by the Gardner is 4950 metres.

Given A.P: 2, 7, 12, …… to 10 terms 

a = 2; d = a₂ – a₁ = 7 – 2 = 5; n = 10 

S_n = \(\frac{n}{2}\) [2a + (n – 1)d] 

∴ S₁₀ = \(\frac{10}{2}\) [2 × 2 + (10 – 1)5] 

= 5 [4 + 9 × 5] 

= 5 [4 + 45]

= 5 × 49 

= 245

Given A.P: -37, -33, -29,…, to 12 terms. 

a = -37; d = a₂ – a₁ = (-33) – (-37) = -33 + 37 = 4; n = 12 

S_n = \(\frac{n}{2}\) [2a + (n – 1)d] 

∴ S₁₂ = \(\frac{12}{2}\) [2 × (-37) + (12 – 1)4] 

= 6 [-74 + 11 × 4] 

= 6 [-74 + 44] 

= 6 × (-30) 

= -180

Let A.P. be a, a + d, a + 2d, a + 3d, .....

Then, First term (a₁)    = a + 0.d

Second term (a₂)       = a + 1.d

Third term (a₃)      = a + 2.d

n^th term (an) = a + (n - 1) d

So, an = a + (n - 1) d is called the n^th term.

The given A.P. is 7, 13, 19, 25,… 

Here, a = 7, d = 13 – 7 = 6 

Since, t_n = a + (n – 1)d

∴ t₁₉ = 7 + (19 – 1)6 

= 7 + 18 × 6 

= 7 + 108 

∴ t₁₉ = 115 

∴ 19^th term of the given A.P. is 115.

The given A.P. is 12, 16, 20, 24,… 

Here, a = 12, d = 16 – 12 = 4 Since, 

t_n = a + (n – 1)d 

∴ t₂₄ = 12 + (24 – 1)4 

= 12 + 23 × 4 

= 12 + 92 

∴ t₂₄ = 104 

∴ 24^th term of the given A.P. is 104.

i. The given sequence is -12, -5,2, 9, 16, 23,30,… 

Here, t₁ = -1, t₂ = -5, t₃ = 2, t₄ = 9 

∴ t₂ – t₁ – 5 – (-12) – 5 + 12 = 7 

t₃ – t₂ = 2 – (-5) = 2 + 5 = 7 

∴ t₄ – t₃ – 9 – 2 = 7 

∴ t₂ – t₁ = t₃ – t₂ = … = 7 = d = constant 

The difference between two consecutive terms is constant. 

∴ The given sequence is an A.P.

ii. t_n = a + (n – 1)d 

∴ t₂₀ = -12 + (20 – 1)7 …[∵a = -12, d = 7] 

= -12 + 19 × 7 

= -12 + 133 

∴ t₂₀ = 121 

∴ 20^th term of the given A.P. is 121.

The correct answer is : (B) is an A.P. Reason d = 4

Option : (C)

Here,

a₇ = 34 and a₁₃ = 64 

i.e.,

a + 6d = 34 ...(i) 

And,

a +12d = 64 ...(ii) 

Subtracting (i) from (ii) we get, 

(12-6) d = 64 - 34 

∴6d = 30 

∴ d = 5 

Substituting d = 5 in (i) we get, 

a+6(5) = 34 

∴a + 30 = 34 

∴ a = 4 

Now,

18^th term is given by a₁₈ = a+17d 

= 4+17×5 

= 89

Given: 9 × (9^th term) = 13 × (13^th term)

To prove: 22^nd term is 0

9 × (a + 8d) = 13 × (a + 12d)

9a + 72d = 13a + 156d

- 4a = 84d

a = - 21d …..Equation 1

Also 22^nd term is given by

a + 21d

Using equation 1 we get

- 21d + 21d = 0

Hence proved 22^nd term is 0.

To Find: The sum of 20 terms of the given AP.

Sum of n terms of an AP with first term a and common difference d is given by

s = n/2[2a+(n -1)d]

Here a = x + y, n = 20, d = - 2y

⇒ S = 10[2x + 2y + 19( - 2y)] = 10[2x + 2y - 38y] = 10[2x - 36y]

⇒ S = 20[x - 18y]

Sum of the series is 20(x - 18y).

Correct answer is B. \(\frac{(b-a)}{(n-1)}\)

‘a’ is the first term of the A.P, 

‘b’ is the (n)^th term and ‘d’ is the common difference of the A.P.

Then, we have b = a + (n–1) d 

= a + (n + 1)d 

Hence, ‘B’ = \(\frac{(b-a)}{(n-1)}\)

Here, 

a_n = xn+ y 

∴ a₁ = x + y 

And a₂ = 2x+y 

So, common difference of an A.P. is given by 

d = a₂ – a₁ 

= (2x+y)-(x+y) 

= x

nth term Tn​ of an AP is given by Tn​=x.n+y

also we know

d=Tn​−Tn−1​

d=x.n+y−(x(n−1)+y)

  =x.n+y−xn+x−y=x

d=x

The given AP is \(\sqrt{2},\sqrt{8},\sqrt{18},......\)

On simplifying the terms, we get:

\(\sqrt{2},2\sqrt{2},3\sqrt{2},.....\)

Here, a = \(\sqrt{2}\) and d = \((2\sqrt{2}-\sqrt{2})=\sqrt{2}\)

∴ Next term, T₄ = a + 3d = \(\sqrt{2}+3\sqrt{2}=4\sqrt{2}=\sqrt{32}\)

The given AP is \(\sqrt{8},\sqrt{18},\sqrt{32},.......\)

On simplifying the terms, we get:

\(2\sqrt{2},3\sqrt{2},4\sqrt{2},.....\)

Here, \(a=2\sqrt{2}\) and \(d=(3\sqrt{2}-2\sqrt{2})=\sqrt{2}\)

∴ Next term T₄ = a + 3d = \(2\sqrt{2}+3\sqrt{2}=5\sqrt{2}=\sqrt{50}\)

We have

T_n = (7 - 4n)

Common difference = T₂ - T₁

T₁ = 7 - 4 x 1 = 3

T₂ = 7 - 4 x 2 = -1

d = -1 - 3 = -4

Hence, the common difference is -4.

We have

T₇ = a + (n - 1)d

\(\Rightarrow\) a + 6d = -4 ........(1)

T₁₃ = a + (n - 1)d

\(\Rightarrow\) a + 12d = -16 ........(2)

On solving (1) and (2), we get

a = 8 and d = -2

Thus, first term = 8 and common difference = -2

∴ The term of the AP are 8,6,4,2,.........

Assuming the first term as a and common difference as d 

To find : the progression 

So, the sum of first 7 terms is given by 

a + a + d + a + 2d + a + 3d…. a + 6d = 10 

7a + 21d = 10…(i)

In the second part it is given that sum of next seven terms is 17 

a + 7d + a + 8d + a + 9d…. a + 13d = 7 

7a + 70d = 7…(ii) 

Solving (i) and (ii) we get 

10 - 21d = 7 - 70d 

3 = - 49d

d = \(-\frac{3}{49}\)

a = \(\frac{79}{49}\)

Hence,

The sequence is given by a, a + d, a + 2d…. which is

\(\frac{79}{49}\),\(\frac{76}{49}\),\(\frac{73}{49}\),....

For a, 7, b, 23, c… to be in AP

it has to satisfy the condition,

a₅ – a₄ = a₄ – a₃ = a₃ – a₂ = a₂ – a₁ = d

Where d is the common difference

7 – a = b – 7 = 23 – b = c – 23 …(1)

Let us equate,

b – 7 = 23 – b

2b = 30

b = 15 (eqn 1)

And,

7 – a = b – 7

From eqn 1

7 – a = 15 – 7

a = – 1

And,

c – 23 = 23 – b

c – 23 = 23 – 15

c – 23 = 8

c = 31

So a = – 1

b = 15

c = 31

Then, we can say that, the sequence – 1, 7, 15, 23, 31 is an AP

We know that,

The first term of an AP = a

And, the common difference = d.

According to the question,

5^th term, a₅ = 19

Using the n^th term formula,

a_n = a + (n – 1)d

We get,

a + 4d = 19

a = 19 – 4d …(1)

Also,

20^th term – 8^th term = 20

a + 19d – (a + 7d) = 20

12d = 20

d = 4/3

Substituting d = 4/3 in equation 1,

We get,

a = 19 – 4(4/3)

a = 41/3

Then, the AP becomes,

41/3, 41/3 + 4/3 , 41/3 + 2(4/3)

41/3, 15, 49/3

Let a be the first term and d be the common difference of an AP.

a₄ = a + (n- 1)d

= a + (4 – 1)d

= a + 3d

Since 4th term of an AP is zero.

a + 3d = 0

or a = -3d ….(1)

Similarly,

a₂₅ = a + 24d = -3d + 24d = 21d …(2)

a₁₁ = a + 10d = -3d + 10d = 7d ….(3)

From (2) and (3), we have

a₂₅ = 3 x a₁₁

Hence proved.