Question

the sum of first 20 terms of ap is 630 and its first term is 3 find 13 term​

Answer 1

Answer:

33

Step-by-step explanation:

Given---> Sum of first 20 terms = 630

First term = 3

To find---> 13th term of AP.

Solution---> Let first term and common difference of AP be a and d.

ATQ, first term = 3

a = 3

Sum of first 20 terms = 630

We know that ,

Sₙ = n / 2 { 2a + ( n - 1 ) d }

=> S₂₀ = 630

=> 20/2 { 2a + ( 20 - 1 ) d } = 630

=> 10 { 2 ( 3 ) + 19d } = 630

=> 6 + 19 d = 630 / 10

=> 6 + 19 d = 63

=> 19 d = 63 - 6

=> 19 d = 57

=> d = 57 / 19

=> d = 3

Now we have to find 13th term ,

aₙ = a + ( n - 1 ) d

=> a₁₃ = 3 + ( 13 - 1 ) (3)

=> a₁₃ = 3 + ( 10 ) ( 3 )

=> a₁₃ = 3 + 30

=> a₁₃ = 33

Answer 2

$\purple{ \huge \mathtt{ \underline{ \fbox{ \: Solution : \: \: \: }}}}$

Given ,

First term of an AP = 3

Sum of first 20 terms of an AP = 630

We know that , the sum of first n number of AP is given by

$\large \sf \fbox{S_{n} = \frac{n}{2} (2a + (n - 1)d)}$

$\sf \hookrightarrow 630 = (\frac{20}{2} ) \bigg(2 ( 3) + (20 - 1)d \bigg )\\ \\ \sf \hookrightarrow \frac{630}{10} = 6 + 19d \\ \\ \sf \hookrightarrow 63 = 6 + 19d \\ \\ \sf \hookrightarrow d = \frac{57}{19} \\ \\\sf \hookrightarrow d = 3$

Thus , the common difference of an AP is 3

Now , the nth term of an AP is given by

$\sf \large{ \fbox{A_{n} = a + (n - 1)d} }$

Substitute the known values , we obtain

$\sf \hookrightarrow A_{13} = 3 + (13 - 1)3 \\ \\\sf \hookrightarrow A_{13} = 3 + 12(3) \\ \\\sf \hookrightarrow A_{13} = 3 + 36 \\ \\\sf \hookrightarrow A_{13} = 39 </p><p>$

Therefore , the 13th term of an AP is 39