Question

If the ratio of sum of p terms and q terms of an A.P. is p^2:q^2, prove that the common difference is twice the first term.

Answer 1

Answer:

It has been proved below that common difference is twice the first term.

Step-by-step explanation:

We are given that the ratio of sum of p terms and q terms of an A.P. is $p^{2} :q^{2}$ .

The formula for sum of n terms of an AP is given by;

                     $S_n = \frac{n}{2}[2a+(n-1)d]$

Given ratio is;

                         ⇒  $\frac{\frac{p}{2} [2a+(p-1)d]}{\frac{q}{2} [2a+(q-1)d]} = \frac{p^{2} }{q^{2} }$

                         ⇒  $\frac{ 2a+(p-1)d}{ 2a+(q-1)d} = \frac{p^{2} }{q^{2} }* \frac{q}{p}$

                         ⇒  $\frac{ 2a+(p-1)d}{ 2a+(q-1)d} = \frac{p }{q }$

                         ⇒  $q*[2a+(p-1)d] = p*[2a+(q-1)d]$

                         ⇒  $2aq + pdq-dq = 2ap+pqd-pd$

                         ⇒ $q[2a-d] = p[2a-d]$

                         ⇒ $p[2a-d] - q[2a-d] = 0$

                         ⇒ $(p - q)(2a-d) = 0$

This means either (p - q) = 0 or (2a - d) = 0

Since, (2a - d) = 0 so d = 2a .

Hence, it is proved that the common difference is twice the first term.