Question

how many terms of the AP: 9,17,25,.......... must be taken so that their sum is 636

Answer 1

Answer:

n=12

Step-by-step explanation:

a=9;d=8:S=636

n/2[2a+(n-1)d]

n/2[2(9)+(n-1)8]

n/2[18+8n-8]

636=n/2=[8n+10]

4n²+5n-636=0

4n²+(53-48)n-636=0

4n²+53n-48n-636=0

4n(n-12)+53(n-12)=0

(n-12) (4n+53)=0

*By Zero Product*

n-12=0

n=12

Answer 2

Given:

An A.P. 9, 17, 25,...

To find:

The number of terms whose sum is 636.

Solution:

As we know that the in an A.P., where a = first term, d = common difference, the sum (s) of 'n' terms of the A.P. is given by:

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

Now,

as given, we have,

an A.P.

9, 17, 25,...

here a = 9,

d = 17 - 9 = 8

Sum of n terms = 636

So,

$636 = \frac{n}{2} [2(9) + (n - 1)8 ]$

$636 = \frac{n}{2} [18 + 8n - 8 ]$

$636 = \frac{n}{2} (10 + 8n)$

On solving the above, we get

$636 = 5 n + 4 {n}^{2}$

${4n}^{2} + 5n - 636 = 0$

This can be written as

${4n}^{2} + 53n - 48n - 636 = 0$

After making factors we get

$n(4n + 53) - 12(4n + 53) = 0$

$(n - 12) (4n + 53) = 0$

$n - 12 = 0 \:and\: 4n + 53 = 0$

$n = 12 \: and \: \frac{ - 53}{4}$

After neglecting n = -53/4, (as terms cannot be in fraction form. Also, it cannot be negative)

We get,

n = 12

Hence, in the given A.P. the number of terms that together make a sum of 636 is 12.