Question

in an ap sum of the first 17 terms is 43 and sum of first 43 terms is 17 then find the sum of first 60 terms is_____​

Answer 1

Answer:

Nth Term in an Arithmetic progression is given by : Nth term=first term + (n-1)d

Here the sum of ist term and 17th term is 40

Therefore a + a+(16)d = 40

2a + 16d = 40

Also ist term + 18 term =43 (given)

a + a + (17)d =43

2a + 17d = 43

Subtract ist equation from ist

D= 3

Answer 2

ANSWER:

Given:

• Sum of first 17 terms = 43
• Sum of first 43 terms = 17

To Find:

• Sum of first 60 terms

Solution:

We are given that,

$:\longrightarrow\sf S_{17}=43$

And,

$:\longrightarrow\sf S_{43}=17$

We know that,

$:\hookrightarrow\sf S_{n}=\dfrac{n}{2}\bigg(2a+(n-1)d\bigg)$

So,

$:\longrightarrow\sf S_{17}=43$

$:\implies\sf S_{17}=\dfrac{17}{2}\bigg(2a+(17-1)d\bigg)$

$:\implies\sf43=\dfrac{17}{2}\bigg(2a+16d\bigg)$

Taking 2 common in the bracket,

$:\implies\sf43=\dfrac{17}{2\!\!\!/}\times2\!\!\!/\bigg(a+8d\bigg)$

So,

$:\implies\sf43=17(a+8d)$

$:\implies\sf43=17a+136d - - - -(1)$

Now,

$:\longrightarrow\sf S_{43}=17$

$:\implies\sf S_{43}=\dfrac{43}{2}\bigg(2a+(43-1)d\bigg)$

$:\implies\sf17=\dfrac{43}{2}\bigg(2a+42d\bigg)$

Taking 2 common in the bracket,

$:\implies\sf17=\dfrac{43}{2\!\!\!/}\times2\!\!\!/\bigg(a+21d\bigg)$

So,

$:\implies\sf17=43(a+21d)$

$:\implies\sf17=43a+903d - - - -(2)$

Now, we have got 2 equations

$:\implies\sf43=17a+136d - - - -(1)$

And,

$:\implies\sf17=43a+903d - - - -(2)$

Multiplying, (1) with 43 and (2) with 17,

$:\implies\sf43\times43=43(17a+136d)$

$:\implies\sf1849=731a+5848d - - - -(3)$

And,

$:\implies\sf17\times17=17(43a+903d)$

$:\implies\sf289=731a+15351d - - - -(4)$

Subtracting (4) from (3),

$:\implies\sf1849-289=731a-731a+5848d-15351d$

$:\implies\sf1560=-9503d$

Hence,

$:\implies\sf d=\dfrac{1560}{-9503}$

Cancelling the numerator and denominator by 13,

$:\implies\bf d=\dfrac{-120}{731}$

Substituting value of d in (1),

$:\implies\sf43=17a+136\left(\dfrac{-120}{931}\right)$

So,

$:\implies\sf43=17a+\dfrac{8\times-120}{43}$

On taking LCM,

$:\implies\sf43=\dfrac{(17\times43)a-960}{43}$

On transposing 43 from RHS to LHS,

$:\implies\sf43\times43=731a-960$

$:\implies\sf1849+960=731a$

So,

$:\implies\sf2809=731a$

Hence,

$:\implies\bf a=\dfrac{2809}{731}$

Now, we need to find sum of first 60 terms,

So,

$:\hookrightarrow\sf S_{n}=\dfrac{n}{2}\bigg(2a+(n-1)d\bigg)$

$:\implies\sf S_{60}=\dfrac{60}{2}\bigg(2\left(\dfrac{2809}{731}\right)+(60-1)\dfrac{-120}{731}\bigg)$

So,

$:\implies\sf S_{60}=30\bigg(\dfrac{5618}{731}+\dfrac{59\times(-120)}{731}\bigg)$

On simplifying,

$:\implies\sf S_{60}=30\bigg(\dfrac{5618}{731}-\dfrac{7080}{731}\bigg)$

$:\implies\sf S_{60}=30\bigg(\dfrac{5618-7080}{731}\bigg)$

$:\implies\sf S_{60}=30\bigg(\dfrac{-1462}{731}\bigg)$

On dividing (-1462) by (731),

$:\implies\sf S_{60}=30\times(-2)$

Therefore,

$:\implies\bf S_{60}=-60$

$\text{\bf{Hence, the Sum of first 60 terms is -60}}$

Formula Used:

$\:\:\:\bullet\:\:\:\sf S_{n}=\dfrac{n}{2}\bigg(2a+(n-1)d\bigg)$