Q. 9

The 17th term of an AP exceeds its 10th term by 7. Find the common difference.

Sum of the first n term is given by (sn) = [2a + (n-1) d]

The sum of first 7 terms S7 = 49 72

S7 = [2×a+(7-1)d]

49 = [2a+6d]

49 = 7[a+3d]

7 = a + 3d 1

The sum of first 17 terms S17 = 289 172

S17 = [2×a+(17-1)d]

289 = [2a+16d]

289 = 17[a+8d]

17 = a + 8d 2

By subtracting both the equations

17 – 7 = a + 8d – (a+3d)

10 = 5d

d =2

common difference (d) = 2

Substituting‘d’ in equation 1

7 = a + 3d

7 = a + 3(2)

a = 1

Sum of the first n term is given by (sn) = [2a + (n-1) d]

= [2×1 + (n-1)2]

= n[ 1+ (n-1)]

= n2

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