Question

If Sum of term of an AP is 295 and first term 7
common diff 5
Then find the value of no​

Answer 1

Answer :

$\large \boxed{\bf{\star\:\:n\:=\:10\:\:\star}}$

Explanation :

Given :–

• $\sf{ S_n \: = \: 295}$
• $\sf {a \: = \: 7}$
• $\sf {d \: = \: 5}$

To Find :–

• The nth Term which make a sum of 295.

Formula Applied :–

• $\boxed{\bf{\star\:\:S_n\:=\:\dfrac{n}{2}\:[2a\:+\:(n\:-\:1)d]\:\:\star}}$

Solution :–

We have, a = 7 , d = 5 and Sn = 295 .

Putting these values in the Formula :

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

$\rightarrow \sf{295\:=\:\dfrac{n}{2}\:[2(7)\:+\:(n\:-\:1)(5)]}$

$\rightarrow \sf{295\:=\:\dfrac{n}{2}\:(14\:+\:5n \: - \: 5)}$

$\rightarrow \sf{295 \: \times \:2 \:=\:n\:(14\:+\:5n \: - \: 5)}$

$\rightarrow \sf{590 \:=\:n\:(9\:+\:5n)}$

$\rightarrow \sf{590 \:= \: 9n\:+\:5n^{2} }$

$\rightarrow \sf{5n^{2} \: + \: 9n \: - \: 590 \: = \: 0 }$

$\rightarrow \sf{5n^{2} \: + \: 59n \: - \: 50n \: - \: 590 \: = \: 0 }$

$\rightarrow \sf{n(5n \: + \: 59) - 10 (5n \: + \: 59 )\: = \: 0 }$

$\rightarrow \sf{(n \: - \: 10 )(5n \: + \: 59 )\: = \: 0 }$

$\rightarrow \sf{n \: = \: 10 \:, \: ( - \dfrac{59}{5}) }$

∵ n cannot be a fraction , in decimal form or negative .

So will simply ignore the negative fraction value.

∴ 10 terms will give a Sum of 295 .

Answer 2

Question:

If Sum of term of an AP is 295 and first term 7, common difference 5 .Then find the no.of terms.

To find:

★ No.of terms ( n ).

Answer:

$No.of\:terms\:\underline{\:\bold{\sf\pink{n\:is\:10}}\:}$

Given:

★ Sum of n term of an A.P. $\underline{ \: S _{n} = 295 \: }$

$\bigstar First \: term \: \underline{ \: a = 7 \: }$

$\bigstar Common \: difference \: \underline{ \: d = 5\: }$

Step-by-step explanation:

We know that,

$\boxed{ S _{n} = \frac{n}{2} \big[2a + (n - 1)d \big]}$

Now substituting the values,

$\implies 295 = \frac{n}{2} \big[2(7) + (n - 1)5 \big]$

$\implies 590 = n \big[14 + 5n - 5 \big]$

$\implies 590 = n \big[9+ 5n \big]$

$\implies 590 = 9n+ 5 {n}^{2}$

$\implies 5 {n}^{2} + 9n - 590 = 0$

Now factorisation, ( you can also use formula method )

$\implies 5 {n}^{2} + 59n -50n - 590 = 0$

$\implies n(5n + 59) - 10(5n + 59)$

$\implies (5n + 59)(n - 10) = 0$

$\implies 5n + 59 = 0 \: \: or \: \: n - 10 = 0$

$\implies n \not= \frac{ - 59}{5} \: \: or \: \: \boxed{n = 10}$

Since, No. of terms will not be in negative.

$\therefore No.of\:terms\:\underline{\:\bold{n\:is\:10}\:}$

More information:

★ Arithmetic Progression :

In a sequence the difference between two consecutive terms are equal it is called Arithmetic Progression .

★ General form :

$a,a+d,a+2d,a+3d,.....$

$\bigstar \: {n}^{th} \: term :$

$t _{n} = a + (n - 1)d$

★ No. of terms :

$n = \big(\frac{l - a}{d} \big) + 1$

★ Common difference :

$d = t _{2} - t _{1}$