If the sum of 7 terms of an A.P. is 49 and that of 17 terms is 289, fi

1.	If the sum of 7 terms of an A.P. is 49 and that of 17 terms is 289, find the sum of n terms.
Answer» Given, Sum of 7 terms of an A.P. is 49 ⟹ S₇ = 49 And, sum of 17 terms of an A.P. is 289 ⟹ S₁₇ = 289 Let the first term of the A.P be a and common difference as d. And, we know that the sum of n terms of an A.P is S_n =\(\frac{ n}{2}\)[2a + (n − 1)d] So, S₇= 49 = \(\frac{7}{2}\)[2a + (7 – 1)d] = \(\frac{7}{2}\) [2a + 6d] = 7[a + 3d] ⟹ 7a + 21d = 49 a + 3d = 7 ….. (i) Similarly, S₁₇= \(\frac{17}{2}\)[2a + (17 – 1)d] = \(\frac{17}{2}\) [2a + 16d] = 17[a + 8d] ⟹ 17[a + 8d] = 289 a + 8d = 17 ….. (ii) Now, subtracting (i) from (ii), we have a + 8d – (a + 3d) = 17 – 7 5d = 10 d = 2 Putting d in (i), we find a a + 3(2) = 7 a = 7 – 6 = 1 So, for the A.P: a = 1 and d = 2 For the sum of n terms is given by, S_n = \(\frac{n}{2}\)[2(1) + (n − 1)(2)] = \(\frac{n}{2}\)[2 + 2n – 2] = \(\frac{n}{2}\)[2n] = n² Therefore, the sum of n terms of the A.P is given by n².

If the sum of 7 terms of an A.P. is 49 and that of 17 terms is 289, find the sum of n terms.

Answer»

Given,

Sum of 7 terms of an A.P. is 49

⟹ S₇ = 49

And, sum of 17 terms of an A.P. is 289

⟹ S₁₇ = 289

Let the first term of the A.P be a and common difference as d.

And, we know that the sum of n terms of an A.P is

S_n =\(\frac{ n}{2}\)[2a + (n − 1)d]

So,

S₇= 49 = \(\frac{7}{2}\)[2a + (7 – 1)d]

= \(\frac{7}{2}\) [2a + 6d]

= 7[a + 3d]

⟹ 7a + 21d = 49

a + 3d = 7 ….. (i)

Similarly,

S₁₇= \(\frac{17}{2}\)[2a + (17 – 1)d]

= \(\frac{17}{2}\) [2a + 16d]

= 17[a + 8d]

⟹ 17[a + 8d] = 289

a + 8d = 17 ….. (ii)

Now, subtracting (i) from (ii), we have

a + 8d – (a + 3d) = 17 – 7

5d = 10

d = 2

Putting d in (i), we find a

a + 3(2) = 7

a = 7 – 6 = 1

So, for the A.P: a = 1 and d = 2

For the sum of n terms is given by,

S_n = \(\frac{n}{2}\)[2(1) + (n − 1)(2)]

= \(\frac{n}{2}\)[2 + 2n – 2]

= \(\frac{n}{2}\)[2n]

= n²

Therefore, the sum of n terms of the A.P is given by n².