If the sum of an A.P. is the same for p as for q terms, shew that its

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

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

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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 $s_p = s_q$