Question

If 12th term of the ap is 213 and sum of its first four term is 24. what is sum of first ten terms?

Answer 1

Answer:

713.715

Step-by-step explanation:

Formula of nth term in A.P. = $a_n=a+(n-1)d$

Substitute n = 12

$a_{12}=a+(12-1)d$

$213=a+11d$  --1

Formula of sum of n terms = $S_n =\frac{n}{2}(2a+(n-1)d)$  --a

So, sum of first four term s:

Substitute n = 4

=$24=\frac{4}{2}(2a+(4-1)d)$

=$24=2(2a+3d)$

=$24=4a+6d$ ---2

Solving 1 and 2

Substitute the value of a from 1 in 2

$24=4(213-11d)+6d$

$24=852-44d+6d$

$24=852-38d$

$828=38d$

$21.789=d$

Substitute the value in 1

$213=a+11(21.789)$

$213=a+239.679$

$-26.679=a$

Now to find sum of first 10 terms

Substitute n = 10 in a

$S_{10} =\frac{10}{2}(2(-26.679)+(9)(21.789))$

$S_{10} =\frac{10}{2}(-53.358+196.101)$

$S_{10} =713.715$

Hence the sum of first 10 terms is 713.715

Answer 2

Given : 12th term of an AP is 213 and the sum of its four term is 24

To Find : what is the sum of its first 10 terms

Solution:

aₙ = a + (n-1)d

Sₙ = (n/2)(2a + (n-1)d)

12th term of an AP is 213

=> a + 11d  = 213

sum of its four term is 24

(4/2)(2a + 3d)  = 24

=> 2a  + 3d  = 12

a + 11d  = 213 => 2a + 22d = 426

2a + 3d = 12

=> 19d  = 414

=> d = 414/19

a + 11d  = 213

=> a = 213 - 11d

=> a = 213 -  11 * 414/19

=> a =  -507/19

Sum of 10th terms =  (10/2)(2a + 9d)

= 5(2( -507/19) + 9(414/19))

=(5/19)2712

= 13560/19

=713.684

sum of its first 10 terms = 13560/19 =713.684

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