If 3+5+7+.....tonterms5+8+11+...to10terms=7, then n is equal to
A
35
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B
36
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C
37
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D
38
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Solution
The correct option is A 35 3+5+7+.....tonterms5+8+11+...to10terms=7Sn=n2(2a+(n−1)d)⇒n2[2×3+(n−1)2]102[2×5+(10−1)×3]=7⇒n[6+2n−2]10[10+27]=7⇒n×2(n+2)10×37=7⇒n2+2n=7×5×37⇒n2+2n=1295⇒n2+2n−1295=0⇒n2+37n−35n−1295=0⇒n(n+37)−35(n+37)=0⇒(n+37)(n−35)=0⇒n+37=0n−35=0∴n=−37orn=35
Hence, a is the correct option.