Question

If the sum of an A.P. is the same for p as for q terms, shew that its sum
for (p + q) terms is zero.​

Answer 1

Answer:

I'm assuming by A.P. you mean arithemtic progression.

The general form of the $n$th term sum of an arithemtic progression is $s_n = \frac{n(a_1+a_n)}{2}$, where $a_1$ is the initial term, and $a_n$ is the $n$th term. Then if the sum is the same for $p$ and $q$ terms, where $p \neq q$, then we have that

$s_p =\frac{p(a_1 + a_p)}{2}=\frac{q(a_1 + a_q)}{2}= s_q$.

Multiplying both sides of the equation by 2 yields

$p(a_1 + a_p) = q(a_1+a_q)$,

expanding then gives

$pa_1+pa_p = qa_1+qa_q$.

Substituting $a_p$ and $a_q$ for the general form of the $n$th term of an arithmetic progression, $a_n = a_1 + (n-1)d$, where $d$ is the difference between consecutive terms, gives

$pa_1+pa_1+pd-d = qa_1 + qa_1 + qd-d$.

Adding d to both sides yields

$2pa_1 + pd = 2qa_1 + qd$,

factorizing results in

$p(2a_1 + d) = q(2a_1 + d)$.

As we know that $p \neq q$, then the above equation holds only if $2a_1+d = 0$. Hence, $2a_1+ d = 0$.

We also notice that  the two sums being equal implies $d = 0$, otherwise the terms of the progression would be of increasing size and the $p$th and $q$th sum could not be equal, which would contradict our assumption that $s_p = s_q$

Hence, we have that $2a_1 = 0$, which clearly implies that $a_1 = 0$.

Then we find the $(p+q)$th sum:

$\frac{(p+q)(a_1 + a_{p+q})}{2}$

We know that $a_{p+q} = a_1 + (p+q)d$. Substituting in the known values for $a_1$ and $d$ gives $a_{p+q} = 0 + (p+q)0 = 0$

substituting this into $\frac{(p+q)(a_1 + a_{p+q})}{2}$ yields

$\frac{(p+q)(0+0)}{2}$

which is indeed equal to zero. Hence we have shown what was required.