Question

$\huge\pink{ \bigstar} \: \bf \color{navy}Que \color{blue}st \purple{ion}$How many terms of the AP -1,-5,-9 must be taken to get a sum of -496​

Answer 1

$\red{\large{\qquad\qquad \underline{ \pmb{{ \mathbb{ \maltese  \:  QUESTION \:   \maltese }}}}}}$

How Many terms of the A.P. 

-1 , -5 , -9 , . . . should be taken to get a Sum of -496.

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$\green{\large{ \qquad \qquad \underline{ \pmb{{ \mathbb{ \maltese  \:  GIVEN \:   \maltese }}}}}}$

An A.P. -1 , -5 , -9 , . . . 

So,

First term = a = -1

And

Common difference = d = -4

Let Required Number of turns be = n

So,

$\bf Sum = S_{n} = - 496$

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$\blue{\large{\qquad \qquad \underline{ \pmb{{ \mathbb{ \maltese  \:  TO \: \: FIND \:   \maltese }}}}}}$

Value of n

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$\green{\large{ \qquad \underline{ \pmb{{ \mathbb{ \maltese  \:  KEY \: \: FORMULA \:   \maltese }}}}}}$

Sum of n terms if an A.P is given by :

$\bigstar \: \: \bf S_{n} = \dfrac{n}{2} \bigg(2a + (n - 1)d \bigg)$

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$\purple{\large{\qquad\qquad \underline{ \pmb{{ \mathbb{ \maltese  \: SOLUTION\:   \maltese }}}}}}$

We have, 

a = -1 

d = -4

Sum = -496

Applying Formula of Sum ,

$\sf - 496 = \dfrac{n}{2} \bigg\{2( - 1) + (n - 1)( - 4) \bigg \} \\ \\ \sf - 496 = \frac{n}{2} \bigg \{2 - 4n \bigg \} \\ \\ \sf- 496 = n(1 - 2n) \\ \\ \sf- 496 = n - 2 {n}^{2} \\ \\ \implies \bf2 {n}^{2} - n - 496 = 0$

$\sf2 {n}^{2} - 32n + 31n - 496 = 0 \\ \\ \sf2n(n - 16) + 31(n - 16) = 0 \\ \\ \sf (2n + 31)(n - 16) = 0 \\ \\ \implies \bf n = 16 \: \: or \: \: n = \frac{ - 31}{2}$

As n is number of terms so it can't be Negative.

Hence ,

$\Large \purple{ \underline{\boxed{ \bf n = 16 }}}$

$\Large \red{\mathfrak{  \text{W}hich \:   \: is  \:  \: the  \:  \: required} }\\ \huge \red{\mathfrak{ \text{ A}nswer.}}$

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Answer 2

Answer:

Given :-

• The AP of - 1, - 5, - 9 must be taken to get a sum of - 496.

To Find :-

• How many numbers of terms can be found in an AP.

Formula Used :-

$\clubsuit$ Sum of first n terms of an AP Formula :

$\mapsto \sf\boxed{\bold{\pink{S_n =\: \dfrac{n}{2}\bigg\lgroup 2a + (n - 1)d\bigg\rgroup}}}$

where,

• $\sf S_n$ = Sum of first n terms of an AP
• a = First term of an AP
• d = Common difference
• n = Number of terms of an AP

Solution :-

Given :

$\bigstar\: \: \: \bf{First\: term\: of\: an\: AP\: (a) =\: - 1}$

$\bigstar\: \: \: \bf{Common\: difference\: (d) =\: - 4}$

$\bigstar\: \: \: \bf{Sum\: of\: n\: terms\: (S_n) =\: - 496}$

According to the question by using the formula we get,

$\implies \sf - 496 =\: \dfrac{n}{2}\bigg\lgroup 2(- 1) + (n - 1)(- 4)\bigg\rgroup$

$\implies \sf - 496 =\: \dfrac{n}{2}\bigg\lgroup -2 + (n - 1)(- 4)\bigg\rgroup$

$\implies \sf - 496 =\: \dfrac{n}{2}\bigg\lgroup - 2 - 4(n - 1)\bigg\rgroup$

$\implies \sf - 496 =\: \dfrac{n}{2}\bigg\lgroup - 2 - 4n + 4\bigg\rgroup$

$\implies \sf - 496 =\: \dfrac{n}{2}\bigg\lgroup - 2 + 4 - 4n\bigg\rgroup$

$\implies \sf - 496 =\: \dfrac{n}{2}\bigg\lgroup 2- 4n\bigg\rgroup$

$\implies \sf - 496 =\: n(1 - 2n)$

$\implies \sf - 496 =\: n - 2n^2$

$\implies \sf - 496 - n + 2n^2 =\: 0$

$\implies \sf 2n^2 - n - 496 =\: 0$

$\implies \sf 2n^2 - (32 - 31)n - 496 =\: 0$

$\implies \sf 2n^2 - 32n + 31n - 496 =\: 0\: \: \bigg\lgroup \small\sf\bold{\pink{By\: splitting\: middle\: term}}\bigg\rgroup$

$\implies \sf 2n(n - 16) + 31(n - 16) =\: 0$

$\implies \sf (2n + 31)(n - 16) =\: 0$

$\implies \bf{(2n + 31) =\: 0}$

$\longrightarrow \sf 2n + 31 =\: 0$

$\longrightarrow \sf 2n =\: - 31$

$\longrightarrow \sf\bold{\purple{n =\: \dfrac{- 31}{2}}}\: \: \bigg\lgroup \small\sf\bold{\pink{Number\: of\: terms\: can't\: be\: negetive (- ve)}}\bigg\rgroup$

$\implies \bf{(n - 16) =\: 0}$

$\longrightarrow \sf n - 16 =\: 0$

$\longrightarrow \sf \bold{\red{n =\: 16}}$

$\therefore$ The number of terms can be found in an AP is 16.

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EXTRA INFORMATION :-

$\clubsuit$ General term (nth term) of an AP Formula :

$\mapsto \sf\boxed{\bold{\pink{a_n =\: a + (n - 1)d}}}$

where,

• $\sf a_n$ = nth term of an AP
• a = First term of an AP
• n = Number of terms of an AP
• d = Common difference