Question

If the sum of first 7 Terms of an AP is 49 { that of 17
terms ej 289 find the sum of first n terms​

Answer 1

Answer:

n^2

Step-by-step explanation:

sum of first seven terms = 49

sum of first seventeen terms ,= 289

Sn = n/2 (2a + (n-1)d)

S7 =

$s7 = \frac{7}{2} (2a + (7 - 1)d) = 49 \\ \frac{7}{2} (2a + 6d) = 49 \\ \frac{7}{2} \times 2(a + 3d) = 49 \\ a + 3d = \frac{49}{7} \\ a + 3d = 7$

S17 =

$s17 = \frac{17}{2} (2a + (17 - 1)d) = 289 \\ \frac{17}{2} (2a + 16d) = 289 \\ \frac{17}{2} \times 2(a + 8d) = 289 \\ a + 8d = \frac{289}{17} \\ a + 8d = 17$

S17 - S7

a + 8d = 17

- a - 3d = -7

5d = 10

d = 10/5 = 2

substitute d = 2 in a + 3d = 7

a + 3(2) = 7

a + 6 = 7

a = 7 - 6

a = 1

a = 1 d = 2

Sn =

$= \frac{n}{2} (2a + (n - 1)d) \\ = \frac{n}{2} (2(1) + (n - 1)(2)) \\ = \frac{n}{2} (2 + 2n - 2) \\ = \frac{n}{2} \times 2n \\ = n \times n \\ = {n}^{2}$

hope you get your answer