Question

how many terms of ap : 9,17,25.........should be taken to give a sum 636.

Answer 1

Heya!!!

Here is your answer.

✴✴✴✴✴✴✴✴✴✴✴✴✴✴

☑ The pattern is 9,17,25.........

☑ The sum of the numbers is 636

If we see the pattern we will see,

↪ (8×1)+1 = 9

↪ (8×2)+1 = 16

↪ (8×3)+1 = 25

✒ 8x + 1 ; here 'x' is the serial no. of the last number or total terms

Now,

♦ We know,
the sum of the numbers of any pattern = {(last number + 1st number)÷2} × serial no. of the last number or total terms.

according to the rule,

636 = [ { ( 8x+ 1 ) + 9 } \ 2 ] × x

• 636 = {8x + 10) \2} × x

• 636 = {(8x\2) + (10\ 2)} × x

• 636 = (4x + 5) × x

• 636 = 4x² + 5x

• 636 = 4x² + 5x

• 636 - 636 = 4x² + 5x - 636

• 0 = 4x² + 5x - 636

• 0 = 4x² + 53x - 48x - 636

• 0 = x ( 4x + 53) - 12 ( 4x + 53)

• (4x + 53) (x - 12) = 0

Now,
if , x - 12 = 0

• x = 12

or, 4x + 53 = 0

• 4x = - 53

• x = - 53 \ 4

But we know serial no. or total terms can not be negative.

So, x = 12

▪ The last number will be {(8×12)+1} = 97

If we want to prove we can write,

(9 + 16 + 25 + 33 + 41 + 49 + 57 + 65 + 73 + 81 + + 89 +97) = 636

❇ So, There are 12 terms in total of the pattern : 9,16,25.....should be taken to give a sum of 636.

❇ Hope it will help you, friend. ☺

Answer 2

$\bf\huge\boxed{\boxed{\bf\huge\:Hello\:Mate}}}$

$\bf\huge Let: first\: term\; be\: a \:and\: CD\: = 17 - 9 = 8$

$\bf\huge => S_{n} = 636$

$\bf\huge => \frac{N}{2}[2a + (n - 1)d] = 636$

$\bf\huge => \frac{N}{2}[2\times 9 + (n - 1)8] = 636$

$\bf\huge => \frac{N}{2} (8n - 10) = 636$

$\bf\huge => n(4n + 5) = 636$

$\bf\huge => 4n^2 + 5n + 636 = 0$

$\bf\huge => n = \frac{-5 + \sqrt{25 - 4\times 4\times -636}}{2\times 4}$

$\bf\huge = \frac{-5 + \sqrt{25 + 10176}}{8}$

$\bf\huge = \frac{- 5 + \sqrt{10201}}{8}$

$\bf\huge = \frac{-5 + 101}{8}$

$\bf\huge = \frac{96}{8} , \frac{-106}{8}$

$\bf\huge = 12 , \frac{-53}{4}$

$\bf\huge But\: n \:cannot\: be\: Negative$

$\bf\huge => n = 12$

$\bf\huge Hence\:Sum\: of\: 12\: terms\: is\: 636$

$\bf\huge\boxed{\boxed{\:Regards=\:Yash\:Raj}}}$