There are two A.P.'s whose common differences differ by unity, but sum of the three consecutive terms in each is 15. If P and P1 be the products of these terms such that P1P=87, then find the common term for all such possible A.P's?

Question

There are two A.P.'s whose common differences differ by unity, but sum of the three consecutive terms in each is 15. If P and P1​ be the products of these terms such that P1​P​=87​, then find the common term for all such possible A.P's?

Filo · Accepted Answer

Let  a−d,a,a+d and A−p,A,A−p are the A.Ps where p=d+1...(1)⇒S=3a=3A=15 ∴a=A=5 Also P1​P​=A(A2−p2)a(a2−d2)​=87​⇒8[25−d2]=7[25−(d+1)2],by(1)or200−175=8d2−7[d2+2d+1]∴d2−14d−32=0 or d=−2,16and hence p=d+1=−1,17. Also a=A=5 Hence the two A.P.'s are given by 7,5,3... and 6,5,4... or ...−11,5,21...−12,5,22... whose three consecutive terms written above the given conditions.Hence common for all A.Ps are 5.

Question Text	There are two A.P.'s whose common differences differ by unity, but sum of the three consecutive terms in each is 15. If P and $P_{1}$ be the products of these terms such that $\frac{P}{P _{1}} = \frac{7}{8},$ then find the common term for all such possible A.P's?
Answer Type	Text solution:1
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There are two A.P.'s whose common differences differ by unity, but sum of the three consecutive terms in each is 15. If P and $P_{1}$ be the products of these terms such that $\frac{P}{P _{1}} = \frac{7}{8},$ then find the common term for all such possible A.P's?

Text solutionVerified

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There are two A.P.'s whose common differences differ by unity, but sum of the three consecutive terms in each is 15. If P and P1​ be the products of these terms such that P1​P​=87​, then find the common term for all such possible A.P's?

Text solutionVerified

Practice questions from similar books

There are two A.P.'s whose common differences differ by unity, but sum of the three consecutive terms in each is 15. If P and $P_{1}$ be the products of these terms such that $\frac{P}{P _{1}} = \frac{7}{8},$ then find the common term for all such possible A.P's?