Questions
The number of terms of the A.P. 1, 4, 7, . . . that must be taken to obtain a sum of 715
is
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a
24
b
23
c
22
d
21
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Correct option is C
Here a = 1, d = 3 and Sn=715 We have715=n2[2a+(n−1)d]=n2[2(1)+(n−1)(3)]⇒ 1430=3n2−n or 3n2−n−1430=0n=16(1±1+3×4×1430)=16(1±131)=22,−65/3As n must be a natural number n = 22Similar Questions
If the sum of first terms of an is , then the sum of squares of these terms is
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