Question

how many terms of Ap :9 ,17, 25,....must be taken to give sum of 636​

Answer 1

Given :

• Sum of the terms of the AP = 636

• Arithmetic Progression (AP) series = 9,17,25,....,n

• First term of the AP = 9

To find :

Number of terms of the AP

Solution :

Common difference of the AP :.

We know the formula for Common Difference i.e,

$\underline{\bf{d = a_{n} - a_{n - 1}}}$

Where :

• $\bf{d}$ = Common Difference

• $\bf{a}$ = Term of the AP

Now using the formula for Common Difference and substituting the values in it, we get :

Here ,

• $\bf{a_{n}}$ = 25
• $\bf{a_{n - 1}}$ = 17

$:\implies \bf{d = a_{n} - a_{n - 1}}$

$:\implies \bf{d = 25 - 17}$

$:\implies \bf{d = 8}$

$\boxed{\therefore \bf{Common\:Difference\:(d) = 8}}$

Hence, the common difference of the AP is 8.

No. of terms of the AP :

We know the formula for sum of n terms of the AP i.e,

$\boxed{\bf{s_{n} = \dfrac{n}{2}\bigg[2a + (n - 1)d\bigg]}}$

Where :-

• s = Sum of the terms of the AP
• n = No. of terms of the AP
• a = First term of the AP
• d = Common Difference

Now , using the formula for sum of terms and substituting the values in it, we get :

$:\implies \bf{s_{n} = \dfrac{n}{2}\bigg[2a + (n - 1)d\bigg]}$

$:\implies \bf{636 = \dfrac{n}{2}\bigg[2 \times 9 + (n - 1)8\bigg]}$

$:\implies \bf{636 = \dfrac{n}{2}\bigg[18 + 8n - 8\bigg]}$

$:\implies \bf{636 = \dfrac{n}{2}\bigg[10 + 8n\bigg]}$

$:\implies \bf{636 = n \times \bigg(\dfrac{10}{2} + \dfrac{8n}{2}\bigg)}$

$:\implies \bf{636 = n \times (5 + 4n)}$

$:\implies \bf{636 = 5n + 4n^{2}}$

$:\implies \bf{0 = 4n^{2} + 5n - 636}$

$:\implies \bf{0 = 4n^{2} + (- 48 + 53)n - 636}$

$:\implies \bf{0 = 4n^{2} - 48n + 53n - 636}$

$:\implies \bf{0 = 4n(n - 12) -+53(n - 12)}$

$:\implies \bf{0 = (n - 12)(4n + 53)}$

$:\implies \bf{0 = (n - 12 = 0) / (4n + 53 = 0)}$

$:\implies \bf{0 = (n = 12) / (4n = -53)}$

$:\implies \bf{0 = n = 12 / n = \dfrac{-53}{4}}$

$\boxed{\therefore \bf{No.\:of\:terms\:(n) = 12,\dfrac{-53}{4}}}$

Since, the no. of terms of an AP can't be negative , the orginal value of n is 12

Hence, no. of terms of the AP is 12.

Answer 2

Given:

✯ A.P. = 9,17,25,.,.,....

✯ Sum of terms = 636

Find:

✮ How many terms will be taken to give the sum as 636

Solution:

Here,

❆ $\red{\sf a = 9}$

❆ $\blue{\sf d = a_2 - a_1}$

$:=\blue{\sf d = 17 - 9}$

$:=\blue{\sf d = 8}$

we, know that

$\boxed{\purple{\sf \to S_n = \dfrac{n}{2}[2a + (n - 1)d]}}$

where,

• $\sf S_n$ = 636
• a = 9
• d = 8

So,

$\pink{\sf \to S_n = \dfrac{n}{2}[2a + (n - 1)d]}$

$\pink{\sf \to 636 = \dfrac{n}{2}[2(9) + (n - 1)(8)]}$

$\pink{\sf \to 636 = \dfrac{n}{2}[18 + 8n - 8]}$

$\pink{\sf \to 636 = \dfrac{n}{2}[10 + 8n]}$

$\pink{\sf \to 636 \times 2 =n[10 + 8n]}$

$\pink{\sf \to 1272 =10n + 8 {n}^{2} }$

$\pink{\sf \to 8 {n}^{2} + 10n - 1272 = 0 }$

$\pink{\sf \to 2(4 {n}^{2} + 5n - 636) = 0 }$

$\pink{\sf \to 2(4 {n}^{2} + 53n - 48n - 636) = 0 }$

$\pink{\sf \to 4 {n}^{2} + 53n - 48n - 636 = \dfrac{0}{2} }$

$\pink{\sf \to n(4n + 53) - 12(4n + 53)= 0}$

$\pink{\sf \to (n - 12)(4n + 53)= 0}$

$\begin{gathered} \sf n - 12 = 0 \\ \sf n = 12 \end{gathered} \qquad \begin{gathered} \sf 4n + 53 = 0 \\ \sf 4n = - 53 \\ \sf n = \dfrac{ - 53}{4}\end{gathered}$

$\underline{\purple{\tiny{ \sf \therefore ignoring \: n = \dfrac{ - 53}{4} \: bcz \: n \: cannot \: be \: in \: negative.}}}$

Hence, number of terms to get Sum as 636 is 12