Question

$\underline{\Huge{\sf Question:-}}$

In an Arithmetic progression

$\sf t_n=3n+2$
Find $\sf S_{61}$​

Answer 1

Required Answer:-

Given:

• Nth term of an A.P. = 3n + 2

To Find:

• Sum of first 61 terms.

Solution:

We have,

➡ Nth term = 3n + 2

Therefore,

➡ First term = 3 × 1 + 2 = 5

➡ Second term = 3 × 2 + 2 = 8

Therefore,

➡ Common Difference = 8 - 5 = 3

A.P. is - 5, 8, 11, 14, 17,. . . . . .

Now, Sum of first n terms of an A.P. is given by the formula,

➡ S = n/2[2a + (n - 1)d]

Therefore, sum of 61 terms will be,

= 61/2 × [2 × 5 + (61 - 1) × 3]

= 61/2 × (10 + 180)

= 61 × 190/2

= 61 × 95

= 5795

★ Hence, the sum of 61 terms of the A.P. will be 5795

Answer:

• Sum of 61 terms of the given A.P. will be 5795.

Learn More:

• A.P. - Stands for Arithmetic Progression, is a sequence in which difference between consecutive terms remains same (constant).
• Example of A.P: 1, 2, 3, 4...

Answer 2

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

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

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

GIVEN :-

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

TO FIND :-

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

FORMULA USED :-

____________________________________________

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

____________________________________________

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.

____________________________________________

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

____________________________________________

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,

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

$\tt \implies (d) = 3$

____________________________________________

Till now we have found :-

➲First term of the A.P = 5

➲Common Difference = 3

____________________________________________

∴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.