In an A.P., if the 5th and 12th terms are 30 and 65 respectively, wh

1.	In an A.P., if the 5th and 12th terms are 30 and 65 respectively, what is the sum of first 20 terms?
Answer» 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.

In an A.P., if the 5th and 12th terms are 30 and 65 respectively, what is the sum of first 20 terms?

Answer»

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.