One of the most important and exam oriented question from arithmetic progression chapter in which we have been asked to find the sum of first n terms if the sum of first 7 terms of an AP is 49 and the sum of its first 17 terms is 289.
Book – RS Aggarwal, Class 10, chapter 5C, question no 29
By using Sn​=2n​[2a+(n−1)d] we have,
S7​=7​/2[2a+(7−1)d]=49
⇒49=7​/2[2a+(7−1)d]
⇒49=7/2​(2a+6d)
⇒7=a+3d
⇒a+3d=7……………….(i)
S17​=17/2​[2a+(17−1)d]=289
⇒289=17​/2[2a+(17−1)d]
⇒289=17/2​(2a+16d)
⇒17=a+8d
⇒a+8d=17………………….(ii)
Substituting (i) from (ii), we get
5d=10 or d=2
From equation (i),
a+3(2)=7
a+6=7 or a=1
Sn​=n/2​[2(1)+(n−1)2]
=n/2​[2+(n−1)2]
=n/2​(2+2n−2)
=n^2