Question

In an ap sum of first 17 term is 43 sum of first 43 terms is 17 then find sum of first 60 terms​

Answer 1

Given:

In an A.P. the sum of the first 17 terms is 43 and the sum of the first 43 terms is 17

To find:

The sum of the first 60 terms​

Solution:

The formula of the sum of n terms of an A.P. is:

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

where Sₙ = sum of n terms, a = first term, d = common difference and n = no. of terms

∴ $S_1_7 = \frac{17}{2} [2a + (17-1)d] = 43$

$\implies \frac{17}{2} [2a + 16d] = 43$

$\implies 17[a + 8d] = 43$

$\implies 17a + 136d = 43$ . . . . (1)

and

∴ $S_4_3 = \frac{43}{2} [2a + (43-1)d] = 17$

$\implies \frac{43}{2} [2a + 42d] = 17$

$\implies 43[a + 21d] = 17$

$\implies 43a + 903d = 17$ . . . . (2)

 

On multiplying equation (1) by 43 and equation (2) by 17, we get

43(17a + 136d) = 43 × 43

⇒ 731a + 5848d = 1849 . . . . (3)

and

17(43a + 903d) = 17 × 17

⇒ 731a + 15351d = 289 . . . . (4)

On subtracting equation (3) and (4), we get

731a + 5848d = 1849  

731a + 15351d = 289

-       -                 -

--------------------------

-9503d =1560

--------------------------

∴ d = $- \frac{1560}{9503} = \bold{- \frac{120}{731}}$

On substituting the value of d in equation (3), we get

$731a + (8\times - 120) = 1849$

$\implies 731a - 960= 1849$

$\implies 731a= 1849 + 960$

$\implies 731a= 2809$

$\implies \bold{a= \frac{2809}{731}}$

Now, we get  

The sum of the first 60 terms is,

= $S_6_0$

= $\frac{60}{2} [2 \times \frac{2809}{731} + (60 -1)(\frac{-120}{731} ) ]$

=  $30 [2 \times \frac{2809}{731} + 59(\frac{-120}{731} ) ]$

= $30 [ \frac{5618}{731} - \frac{7080}{731} ]$

= $30 [- \frac{1462}{731} ]$

= $30\times -2$

= $\bold{-60}$

Thus, the sum of the first 60 terms is → -60.

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