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mario62 [17]
3 years ago
6

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.

Mathematics
2 answers:
tankabanditka [31]3 years ago
7 0

\sf\underline{\underline{Question:}}

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.

$\sf\underline{\underline{Solution:}}$

  • $\sf\bold\purple{||100||}$

$\space$

$\sf\underline\bold\red{||Step-by-Step||}$

$\sf\bold{Given:}$

  • $\sf\bold{S6=36}$
  • $\sf\bold{S16=255}$

$\space$

$\sf\bold{To\:find:}$

  • $\sf\bold{The \: sum\:of\:the\:first\:ten\:numbers}$

$\space$

$\sf\bold{Formula\:we\:are\:using:}$

$\implies$ $\sf{ Sn=}$ $\sf\dfrac{N}{2}$ $\sf\small{[2a+(n-1)d]}$

$\space$

$\sf\bold{Substituting\:the\:values:}$

→ $\sf{S6=}$ $\sf\dfrac{6}{2}$ $\sf\small{[2a+(6-1)d]}$

→ $\sf{36 = 3[2a+(6-1)d]}$

→$\sf{12=[2a+5d]}$ $\sf\bold\purple{(First \: equation)}$

$\space$

$\sf\bold{Again,Substituting \: the\:values:}$

→ $\sf{S16}$ $\sf\dfrac{16}{2}$ $\sf\small{[2a+(16-1)d]}$

→ $\sf{255=8[2a + (16-1)d]}$

:: $\sf\dfrac{255}{8}$ $\sf\small{=31.89=32}$

→ $\sf{32=[2a+15d]}$ $\sf\bold\purple{(Second\:equation)}$

$\space$

$\sf\bold{Now,Solve \: equation \: 1 \:and \:2:}$

→ $\sf{10=20}$

→ $\sf{d=}$ $\sf\dfrac{20}{10}$ $\sf{=2}$

$\space$

$\sf\bold{Putting \: d=2\: in \:equation - 1:}$

→ $\sf{12=2a+5\times 2}$

→ $\sf{a = 1}$

$\space$

$\sf\bold{All\:of\:the\:above\:eq\: In \: S10\:formula:}$

$\mapsto$ $\sf{S10=}$ $\sf\dfrac{10}{2}$ $\sf\small{[2\times1+(10-1)d]}$

$\mapsto$ $\sf{5(2\times1+9\times2)}$

$\mapsto$ $\sf\bold\purple{5(2+18)=100}$

$\space$

$\sf\small\red{||Hence , the \: sum\: of \: the \: first\:10\: terms\: is\:100||}$

_____________________________

Cloud [144]3 years ago
4 0

Answer:

100

Step-by-step explanation:

We have the sum of first n terms of an AP,

Sn = n/2 [2a+(n−1)d]

Given,

36= 6/2 [2a+(6−1)d]

12=2a+5d ---------(1)

256= 16/2 [2a+(16−1)d]

32=2a+15d ---------(2)

Subtracting, (1) from (2)

32−12=2a+15d−(2a+5d)

20=10d ⟹d=2

Substituting for d in (1),

12=2a+5(2)=2(a+5)

6=a+5 ⟹a=1

∴ The sum of first 10 terms of an AP,

S10 = 10/2 [2(1)+(10−1)2]

S10 =5[2+18]

S10 =100

This is the sum of the first 10 terms.

Hope it will help.

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