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

1.	If the sum of 7 terms of an A.P. is 49 and that of 17 term is 289, find the sum of n terms.
Answer» S₇= 49 \(\frac{7}{2}\)[2a + 6d] = 49 a + 3d = 7 (i) S₁₇= 289 \(\frac{17}{2}[2a + 16d] =289\) a + 8d = 17 (ii) Subtract (i) from (ii), we get 5d = 10 d = 2 Put d = 2 in (i), we get a = 7 – 6 = 1 Sn= \(\frac{n}{2}\)[2(1) + (n – 1)2] = n [1 + n – 1] = n²

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

Answer»

S₇= 49

\(\frac{7}{2}\)[2a + 6d] = 49

a + 3d = 7 (i)

S₁₇= 289

\(\frac{17}{2}[2a + 16d] =289\)

a + 8d = 17 (ii) Subtract

(i) from (ii),

we get

5d = 10

d = 2

Put d = 2 in (i),

we get a = 7 – 6 = 1

Sn= \(\frac{n}{2}\)[2(1) + (n – 1)2]

= n [1 + n – 1]

= n²