Question

if the sum of first 7 terms of an ap is 49 and that of 17th term is289 find the sum of n term

Answer 1

Given that,

S7 = 49
S17 = 289

S7 = 7/2 [2a + (n - 1)d]
S7 = 7/2 [2a + (7 - 1)d]
49 = 7/2 [2a + 16d]
7 = (a + 3d)
a + 3d = 7 ... (i)

Similarly,
S17 = 17/2 [2a + (17 - 1)d]
289 = 17/2 (2a + 16d)
17 = (a + 8d)
a + 8d = 17 ... (ii)
Subtracting equation (i) from equation (ii),
5d = 10
d = 2
From equation (i),
a + 3(2) = 7
a + 6 = 7
a = 1

Sn = n/2 [2a + (n - 1)d]
= n/2 [2(1) + (n - 1) × 2]
= n/2 (2 + 2n - 2)
= n/2 (2n)
= n2

Answer 2

Answer:

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 Sn =n2n2[2a + (n − 1)d] So, S7 = 49 = 7272[2a + (7 – 1)d] = 7272 [2a + 6d] = 7[a + 3d] ⟹ 7a + 21d = 49 a + 3d = 7 ….. (i) Similarly, S17 = 172172[2a + (17 – 1)d] = 172172 [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, Sn  = n2n2[2(1) + (n − 1)(2)] = n2n2[2 + 2n – 2] = n2n2[2n] = n2