If sum of first 6 digits of AP is 36 and that of the first 16 terms is 255,then find the sum of first ten terms.
2 answers:
If sum of first 6 digits of AP is 36 and that of the first 16 terms is 255,then find the sum of first ten terms.
$\sf\underline{\underline{Solution:}}$
- $\sf\bold\purple{||100||}$
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$\sf\underline\bold\red{||Step-by-Step||}$
$\sf\bold{Given:}$
- $\sf\bold{S6=36}$
- $\sf\bold{S16=255}$
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$\sf\bold{To\:find:}$
- $\sf\bold{The \: sum\:of\:the\:first\:ten\:numbers}$
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$\sf\bold{Formula\:we\:are\:using:}$
$\implies$ $\sf{ Sn=}$ $\sf\dfrac{N}{2}$ $\sf\small{[2a+(n-1)d]}$
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$\sf\bold{Substituting\:the\:values:}$
→ $\sf{S6=}$ $\sf\dfrac{6}{2}$ $\sf\small{[2a+(6-1)d]}$
→ $\sf{36 = 3[2a+(6-1)d]}$
→$\sf{12=[2a+5d]}$ $\sf\bold\purple{(First \: equation)}$
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$\sf\bold{Again,Substituting \: the\:values:}$
→ $\sf{S16}$ $\sf\dfrac{16}{2}$ $\sf\small{[2a+(16-1)d]}$
→ $\sf{255=8[2a + (16-1)d]}$
:: $\sf\dfrac{255}{8}$ $\sf\small{=31.89=32}$
→ $\sf{32=[2a+15d]}$ $\sf\bold\purple{(Second\:equation)}$
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$\sf\bold{Now,Solve \: equation \: 1 \:and \:2:}$
→ $\sf{10=20}$
→ $\sf{d=}$ $\sf\dfrac{20}{10}$ $\sf{=2}$
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$\sf\bold{Putting \: d=2\: in \:equation - 1:}$
→ $\sf{12=2a+5\times 2}$
→ $\sf{a = 1}$
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$\sf\bold{All\:of\:the\:above\:eq\: In \: S10\:formula:}$
$\mapsto$ $\sf{S10=}$ $\sf\dfrac{10}{2}$ $\sf\small{[2\times1+(10-1)d]}$
$\mapsto$ $\sf{5(2\times1+9\times2)}$
$\mapsto$ $\sf\bold\purple{5(2+18)=100}$
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$\sf\small\red{||Hence , the \: sum\: of \: the \: first\:10\: terms\: is\:100||}$
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