Question

In a AP In the 12 term is 43 and the sum of its first four terms is 24. Find the sum of its first ten terms​

Answer 1

Step-by-step explanation:

bhai is oues. mein 2 eq. banege

1st-. 43= a+(12-1)d

43= a+ 11d -------1eq.

2nd- 24= 4/2(2a+ 3d)

12= 2a+ 3d ------2eq.

by elimination method,

now solve it

Answer 2

$S_{10}=\frac{3360}{19}$

Step-by-step explanation:

Given:

$a_{12}=43$

$S_4=24$

Let a be the first term and d be the common difference of A.P

$a_n=a+(n-1)d$

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

Using the formula

$a_{12}=a+11d$

$a+11d=43$..(1)

$S_4=\frac{4}{2}(2a+3d$

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

$2a+3d=\frac{24}{2}=12$...(2)

Equation (1) multiply by 2 and then subtract from equation (2)

$-19d=-74$

$d=\frac{74}{19}$

Substitute the value of d in equation (1)

$a+11\times \frac{74}{19}=43$

$a+\frac{814}{19}=43$

$a=43-\frac{814}{19}$

$a=\frac{43\times 19-814}{19}$

$a=\frac{3}{19}$

n=10

$S_{10}=\frac{10}{2}(2\times \frac{3}{19}+9\times \frac{74}{19})$

$S_{10}=5\times (\frac{6+666}{19})$

$S_{10}=5\times\frac{672}{19}=\frac{3360}{19}$

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