Question

If 12th term of an AP is 37 and common difference is

3,find it’s first term and sum of first 12 terms.​

Answer 1

Answer:

• First term of the A. P. = 4
• Sum of first 12th term is 246

Explanation:

Given,

12th term of an A. P. is 37

Whose common difference is 3.

We know,

nth term of an A. P. :-

$\sf \longrightarrow a_n = a + (n - 1)d$

Where, a denotes the first term, n denotes the number of terms and d is the common difference.

By the given data,

$\sf \implies \: 37 = a + (12 - 1)(3)$

$\sf \implies \: 37 = a + (11)(3)$

$\sf \implies \: 37 = a + 33$

$\sf \implies \: 37 - 33 = a$

$\sf \implies \: 4 = a$

∴ First term of the A. P. is 4.

Now, we need to find the sum of first 12 terms.

Sum of first 12 terms is given by :-

$\sf \longrightarrow \: \dfrac{n}{2} [2a + (n - 1)d]$

Where,

• a(first term) = 4
• n(number of terms) = 12
• d(common difference) = 3

Substituting the values,

$\sf \longrightarrow \: \dfrac{12}{2} [2(4) + (12 - 1)(3)]$

$\sf \longrightarrow \: 6 [8+ (11)(3)]$

$\sf \longrightarrow \: 6 [8+ 33]$

$\sf \longrightarrow \: 6 [41]$

$\sf \longrightarrow \: 246$

∴ Sum of first 12 terms is 246

Answer 2

Answer:

The first term = 4

The sum of first 12 terms = 246

Step-by-step explanation:

Given :

12th term of an AP = 37

common difference, d =  3

To find :

it's first term and sum of first 12 terms.​

Solution :

nth term of an A.P is given by,

$\longmapsto \sf a_n=a+(n-1)d$

12th term = 37

a₁₂ = 37

a + (12 - 1)d = 37

a + 11d = 37

a + 11(3) = 37

a + 33 = 37

a = 37 - 33

a = 4

∴ The first term = 4

Sum of first n terms is given by,

$\longmapsto \sf S_n=\dfrac{n}{2}[2a+(n-1)d]$

We have to find the sum of first 12 terms, put n = 12

$\sf S_{12}=\dfrac{12}{2}[2(4)+(12-1)(3)] \\\\ \sf S_{12}=6[8+11(3)] \\\\ \sf S_{12} =6[8+33] \\\\ \sf S_{12} = 6[41] \\\\ \sf S_{12}=246$

Therefore, the sum of first 12 terms of the given A.P is 246