Question

how many terms are there in the series 1+4+7+...,...,...,+64?

Answer 1

Answer:

$here \\ \: \: \: a=1 \\ d= 3 \\ l = 64 \\ \\ l = a + (n - 1)d \\ = 1 + (n - 1)3 \\ = > 64 - 1 = (n - 1)3 \\ = > \frac{63}{3} = n - 1 \\ = > 21 = n - 1 \\ = > n = 21 + 1 = 22 \\ \\ \\ \\ hence \: 22 \: terms \: are \\ \: there \: in \: the \: given \\ \: arithematic \: sequnce \: .$

Answer 2

${\huge{\star{\underbrace{\tt{\red{Answer}}}}}}{\huge{\star}}$

$\Large{\blue{\tt{\underline{\underline{Given \ :-}}}}}$

${\sf{❥ \ Series \ = \ 1+4+7......+64}}$

$\Large{\green{\tt{\underline{\underline{To \ find \ :-}}}}}$

${\sf{✿ \ Number \ of \ terms \ in \ the \ series}}$

$\Large{\orange{\tt{\underline{\underline{Solution \ :-}}}}}$

${\sf{ \hookrightarrow \ First \ term \ = \ 1}}$

${\sf{\hookrightarrow \ Common \ difference \ = \ a_{2} \ - \ a_{1}}}$

${\sf{\hookrightarrow \ Common \ difference \ = \ 4 \ - \ 1 \ ⟹ \ 3}}$

${\sf{\hookrightarrow \ Last \ term \ (a_{n}) \ = \ 64}}$

${\boxed{\pink{\sf \ ✯ \ a_{n} \ = \ a_{1}+(n-1)d}}}$

${\tt{Substituting \ the \ values}}$

${\sf{\leadsto \ 64 \ = \ 1+(n-1)3}}$

${\sf{\leadsto \ 64-1 \ = \ 3n-3}}$

${\sf{\leadsto \ 63+3 \ = \ 3n}}$

${\sf{\leadsto \ \frac{\cancel{66}}{\cancel 3} \ = \ n}}$

${\sf{\leadsto \ n \ = \ 22}}$

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