Question

1) For an A.P, show that $t_m+t_{2n+m}=2t_{m+n}$

2) If the sum of p terms of an A.P is q and the sum of its q terms is p, then find the sum Of its $(p+q)$ term.

Answer 1

$1)$

$\therefore t_m=a+(m-1)d$

$t_{2n+m}=a+(2n+m-1)d$

${2t}_{m+n}=2[a+(n+m-1\left)d\right]$

Adding both,

$t_m+t_{2n+m}$ $=a+(m-1)d+a+(2n+m-1),d=2a+(2n+m-1+m-1)d,=2a+(2n+2m-2)d,=2[a+(n+m-1\left)d\right].\text{(Proved)}$

$2)$

$S_p=q,$ $S_q=p$

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

Subtracting both.

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

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

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

Dividing $(p-q)$ from both sides.

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

$-2=2a+(p+q-1)d$

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

$S_{p+q}=\frac{p+q}{2}\{-2\}=-(p+q)$