Question

How many terms of AP minus 1 minus 5 minus 9 must be taken to get a sum - 496

Answer 1

Answer:

n = 16 or -31/2

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Answer 2

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

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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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 of n terms if an A.P

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

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

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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