Question

how mamy terms of an AP 7,11,15,19,23,... must be taken to get the sum of 250​

Answer 1

Answer:

Let the number of terms be n

$sum \: ofn \: numbers = \frac{n}{2} \times {2a + (n - 1)d}$

2

$250 = \frac{n}{2} \times 2 \times 7 + (n - 1)4 \\ \\ 500 = n(14 + 4n - 4) \\ \\ 500 = 4 {n}^{2} + 10 \\ \\ 500 = {4n}^{2} + 10n \\ \\ {4n}^{2} + 10n - 500 = 0 \\ \\ 4 {n}^{2} + 50n - 40n - 500 = 0 \\ \\ 2n(2n + 25) - 20(2n + 25) = 0 \\ \\ 2n - 20 = 0 \\ 2n + 25 = 0 \\ n = 10 \: \: or \: \: n \: = - \frac{25}{2}$

Since number of terms cant be negative n = 10

Answer 2

Given :

• Sequence ➡ 7,11,15,19,23

• Sum of Ap=250

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To Find :

• How many terms of an Ap=?

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Formulas :

•For Finding Sum of Ap:

$\\ \red\bigstar\boxed{\tt{S_{n}=\dfrac{n}{2}\bigg(2a+(n-1)d\bigg) }}$

•For Finding number of terms :

$\\ \red\bigstar\boxed{\tt{t_{n}=a+(n-1)d}}$

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Solution :

In a AP,

• First term(a)=7

•common difference (d) =11-7=4

•Sum of Ap=250

$\\ \implies\sf{S_{n}=\dfrac{n}{2}\bigg\{ 2a+(n-1)d\bigg\}}$

$\\ \implies\sf{S_{n}=250}$

$\\ \implies\sf{250=\dfrac{n}{2}\bigg\{2(7)+(n-1)4\bigg\}}$

$\\ \implies\sf{250 \times 2=n\bigg(14+4n-4\bigg)}$

$\\ \implies\sf{500=n\bigg(10+4n\bigg)}$

$\\ \implies\sf{500=10n+4n^{2}}$

$\\ \implies\sf{500-10n-4n^{2}}$

$\\ \implies\sf{-4n^{2}-10n+500=0}$

$\\ \implies\sf{4n^{2}+10n-500=0}$

$\\ \implies\sf{2n^{2}+5n-250=0}$

$\\ \implies\underline{\tt{By \: using \: Formula \: method :}}$

$\\ \implies \sf \: x = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a} \\ \\ \\ \implies \sf \: x = \frac{ - 5 \pm \sqrt{25 - 4(2)( - 250)} }{2 \times 2} \\ \\ \\ \implies \sf \: x = \frac{ - 5 \pm \sqrt{2025} }{4} \\ \\ \\ \implies \sf \: x = \frac{ - 5 \pm45}{4} \\ \\ \\ \implies \sf \: x = \frac{ - 5 + 45}{4} \: \: \: \quad \: \: x = \frac{ - 5 - 45}{4} \\ \\ \\ \implies \sf \: x = \frac{40}{4} \: \: \quad \: \: \: \: x = \frac{ - 50}{4} \\ \\ \\ \implies \sf \: \boxed{ \sf{x = 10}} \: \: \quad \: \: \boxed{ \sf{x = \frac{ - 25}{2} }}$

•As we know that,

$\\ \bf{\pink\bigstar \: Number \: of \: terms \: cannot \: be \: negative }$

$\\ \rm{So, \: \dfrac{-25}{2} \: is \: neglected }$

$\therefore \underline{\tt { the \: number \: of \: terms \: is \boxed{10}}}$