Q. 9
The 17th term of an AP exceeds its 10th term by 7. Find the common difference.
Answer :
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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Show that a1, a2,…..an form an AP where is defined as below :
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