Question

Question:-
$In \: an \: Arithmetic \: progression \ \textless \ br /\ \textgreater \ \\ \ \textless \ br /\ \textgreater \ \sf t_n=3n+2  \\ \ \textless \ br /\ \textgreater \ Find \sf S_{61}\ \textless \ br /\ \textgreater$
answer only if u know...fast..plz​

Answer 1

$\huge\purple{\mid{\underline{\overline{\texttt{Question}}}\mid}}$

In an Arithmetic progression

$\tt \implies \: t_n=3n+2$

$\tt \implies \: find \: = s_{61}$

$\huge\pink{\mid{\fbox{\tt{Answer↴}}\mid}}$

5795 is the sum of 61 terms of the A.P

$\huge\red{\mid{\fbox{\tt{Solution}}\mid}}$

GIVEN:-

$\tt \implies \: t_{n} \: = 3n + 2$

TO FIND:-

The sum of first 61 terms $\tt ( \: s_{61})$ ?

FORMULA USED:-

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$\tt \implies \: s_{ {n}^{th} } = \frac{n}{2} [2a + (n - 1)d]$

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where,

➙$\tt ( s_{n ^{th} })$ = sum of $\tt \: {n}^{th}$ of the A.P.

➙n = term number of the number in an A.P

➙a = the First term of the A.P

➙d = common difference between the terms.

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$\tt \implies \: common \: difference(d) \: = t_{2} - t_{1}$

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where,

➙d = common difference between the terms.

➙$\tt \: t_{2} \: =$ second term of the A.P

➙$\tt t_{1} =$ First term of the A.P

CALCULATION:-

As given above ,

$\tt \: t_n=3n+2$

taking n = 1 , then the first term will be :-

$\tt \implies \: t_1=3n+2$

$\tt \implies \: t_1=3(1)+2$

$\tt \implies \: t_1=3+2$

$\tt \implies \: t_1=5$

taking n = 2 ,then the second term will be:-

$\tt \implies \: t_2=3n+2$

$\tt \implies \: t_2=3(2)+2$

$\tt \implies \: t_2=6+2$

$\tt \implies \: t_2=8$

∴The common difference will be :-

$\tt \implies (d) = t_2 - t_1$

substituting the values,

$\tt \implies (d) = 8 - 5$

$\tt \implies (d) = 3$

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Till now we have found :-

➲First term of the A.P = 5

➲Common Difference = 3

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∴sum of the 61 terms of A.P will be :-

$\tt \implies \: s_{ {n}^{th} } = \frac{n}{2} [2a + (n - 1)d]$

$\tt \implies \: s_{ 61} = \frac{n}{2} [2a + (n - 1)d]$

substituting the values,

$\tt \implies \: s_{ 61} = \frac{61}{2} [2(5) + (61 - 1)3]$

$\tt \implies \: s_{ 61} = \frac{61}{2} [10 + (60)3]$

$\tt \implies \: s_{ 61} = \frac{61}{2} (10 + 180)$

$\tt \implies \: s_{ 61} = \frac{61}{2} \times ( 190)$

[Note :- 190 is being divided by 2]

$\tt \implies \: s_{ 61} = 61 \times 95$

$\tt \implies \: s_{ 61} = 5795$

Hence, the sum of 61 terms of the A.P is 5795.