Question

How many terms of the A.P. 43, 39, 35,be taken so that their sum is
252 ?​

Answer 1

Answer:

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Step-by-step explanation:

Given first term(a) = 43

Common difference(d) = 39 - 43 = -4

Sum upto n terms of an AP = n/2 {2a + (n-1)d}

where n is no of terms

=> n { 2a + (n-1)d } /2 = 252

=> n { 2(43) + (n-1)(-4) } = 504

=> n { 86 - 4n + 4 } = 504

=> 86n - 4n^2 + 4n = 504

=> -4n^2 + 90n - 504 = 0

=> -4n^2 - 42n - 48n - 504 = 0

=> -2n ( 2n + 21) -24 ( 2n + 21 ) = 0

=> ( 2n + 21 )( -2n - 24) = 0

=> n = -21/2 (or) n = 12

∴ No of terms are 12 because no of terms can't be in decimals

Answer 2

Answer:

12 is the term for which the sum of an given A.P is 252.

Step-by-step explanation:

As we know that:

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

Given:

A.P= 43,39,35,...

s(n)= 252

a=43

d=39-43= -4

n=?

Now,

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

$252 = \frac{n}{2} (2(43) + (n - 1)( - 4))$

$252 = \frac{n}{2} (86 + (n - 1)( - 4))$

$252 = \frac{n}{2} (86 + ( - 4n + 4))$

$252 = \frac{n}{2} (86 - 4n + 4)$

$252 = \frac{n}{2} (90 - 4n)$

$\frac{252}{(90 - 4n)} = \frac{n}{2}$

$n(90 - 4n) = 252 \times 2$

$90n - 4 {n}^{2} = 504$

$- 4 {n}^{2} + 90n - 504 = 0$

$- 4 {n}^{2} + 48n + 42n - 504 = 0$

$4n(n - 12) + 42(n - 12) = 0$

$( 4n + 42)(n - 12) = 0$

$now$

$4n + 42 = 0$

$4n = - 42$

$n = \frac{ - 42}{4}$

$then$

$n- 12 = 0$

$n= 12$

since, -42/4 cannot gives an exact value then the value is 12 .