Question

If pth terem of an AP is 1/q and qth term 1/p
prove that the sum of the first
'Pq' terms is 2 (pq+1)
-te
P​

Answer 1

Answer:

AnswEr:

$\begin{gathered}\begin{gathered}\frak{Given}\begin{cases} \sf{First\:Term\:(a)\:=\:17}\\ \sf{Last\:Term\:(L)\;=\; 350} \\ \sf{Common\: Difference\:(d)\:=\; 9}\end{cases}\end{gathered}\end{gathered}$

Given

⎩

⎪

⎪

⎨

⎪

⎪

⎧

FirstTerm(a)=17

LastTerm(L)=350

CommonDifference(d)=9

$\:\:\:\:\;\:\;\footnotesize\bold{\underline{\underline{\sf{\red{Formula\:For\:the\: Last\:Term\:is \:-}}}}}$

FormulaFortheLastTermis−

$\dag\;\:\large{\underline{\boxed{\sf{\purple{l\:=\; a +(n -1)d}}}}}$

l=a+(n−1)d

†

$\large\bold{\underline{\sf \green{Putting\:Values-}}}$

PuttingValues−

$:\implies\sf \: 350 = 17 + (n -1)9::⟹350=17+(n−1)9:$

$\implies\sf \:\cancel\dfrac{333}{9}$= (n -1)⟹

9

333

=(n−1)

$:\implies\sf \: (n -1)$ = 37:⟹(n−1)=37:⟹(n−1)=37:⟹(n−1)=37

:

$\implies\sf \: n$ = 37 + 1:⟹n=37+1⟹n=37+1:⟹n=37+1

$:\implies\large\boxed{\sf{\purple{n\:=\;38}}}$::⟹

n=38

:

$\small\bold{\underline{\sf{\pink{Therefore\:Given\:AP\: Contains\: 38\:Terms.}}}}$

ThereforeGivenAPContains38Terms.

$\large\bold{\underline{\sf{Now\: Finding\: Sum -}}}$ NowFindingSum−

$:\implies\sf \: S_{n}$ = \dfrac{n}{2} (a +l)::⟹S

$:\implies\sf \: S_{38}$ = $\dfrac{38}{2}(17 + 350):$⟹S

$:\implies\sf \: 19 \times 367::$⟹19×367:

$:\implies\large\boxed{\sf{\purple{6973}}}$::⟹

$\small\bold{\underline{\sf{\red{Sum \:of \: the \:Terms\: of \: the \: Given\: AP\; is \: 6973.}}}}$