Question
Question: Let \( {{a}_{1}},{{a}_{2}},{{a}_{3}},.... \) be terms of an A.P. If \( \dfrac{\left( {{a}_{1}}+{{a}_...
Let a1,a2,a3,.... be terms of an A.P. If (a1+a2+a3...aq)(a1+a2+a3...ap)=q2p2 , p not equal to q, then a21a6 equals
a. 27
b. 72
c. 4111
d. 1141
Solution
To deal with the type of question we will use the formula of sum of n terms of A.P. and finding the nth term, which is 2n(2a+(n−1)d) and (a+(n−1)d) respectively. In which ‘n’ is the number of terms, ‘a’ is the first term and ‘d’ is the difference of an A.P.
Complete step by step answer:
Moving ahead with the question in step wise manner, we had to find the value of a21a6 so as we know that an means nth term of an A.P. means we can say that a6 and a21 are 6th and 21th term of an A.P. and as we also know that nth term of an A.P. is (a+(n−1)d) , so we can say that 6th and 21th term will be
a1+(6−1)da1+5d and a1+(21−1)da1+20d
So from here we can say that we need to find out the value of a21a6 which can be written asa1+20da1+5d, means we if we get the value of ‘a’ and ‘d’ or relation among them then we will get the value. So let us use the given condition to find the relation or value of ‘a’ and ‘d’.
So we have (a1+a2+a3...aq)(a1+a2+a3...ap)=q2p2 , in LHS side in numerator and denominator it represents the sum of terms starting from 1st term to pth and qth term, which we can write it as, sum of terms of A.P. starting from 1st term to pth is 2p(2a1+(p−1)d) similarly we can write for denominator series, which will be2q(2a1+(q−1)d). So we can replace the series with this formula, which will give us;
2q(2a1+(q−1)d)2p(2a1+(p−1)d)=q2p2
On further simplifying it we will get;
q(2a1+(q−1)d)p(2a1+(p−1)d)=q2p22a1+(q−1)d2a1+(p−1)d=qp
By cross multiplying we will get;
2a1q+qpd−qd=2a1p+qpd−pd2a1q+qpd−qd−2a1p−qpd+pd=02a1(q−p)−(q−p)d=0
By taking the q−p common we will get;
(q−p)(2a1−d)=0
So from here we can say that either (q−p) is equal to zero, or (2a1−d) equal to zero. So by putting (2a1−d) equal to zero we will get;
2a1−d=02a1=d
So we got relation i.e. 2a1=d
Now put the value of ‘d’ in the expression whose we need to find the value which is;