Question

How many terms of the series -8-4+0+......... must be taken
so that the sum be 132?​

Answer 1

Answer:

hope you got your answer

Answer 2

$\huge\mathfrak\blue{Answer:}$

Arithmetic Progression:

• An Arithmetic Progression is the series of number that increase or decrease by same amount each time
• For ex : 2 , 4 , 6 , 8

Given:

• We have been given a series of number such that -8 , -4 , 0 , 4 ...
• Sum of the series to be 132

To Find:

• We have to find the number of terms so that sum should be equal to 132

Solution:

We have been given a series of number

$\text{-8 , -4 , 0 , 4 , 8}$

On observing the series we found that it is an Arithmetic Progression

$\boxed{\sf{First \: Term \: (a) = - 8}}$

$\boxed{\sf{Common \: Difference \: (d) = 8-4 = 4}}$

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$\underline{\large\mathfrak\red{According \: to \: the \: Question:}}$

$\large{\boxed{\sf{Sum \: of \: Series = 132}}}$

Sum of series in AP is given by :

$\implies \sf{ \dfrac{n}{2} \: [ \: 2a + (n-1)d \: ] = 132}$

$\implies \sf{n \: [ \: 2a + (n-1)d \: ] = 132 \times 2}$

Substituting the Values

$\implies \sf{n \: [ \: 2(-8)+ (n-1)4 \: ] = 264}$

$\implies \sf{n \: [ \: -16 + 4n - 4 \: ] = 264}$

$\implies \sf{n \: [ \: 4n - 20 \: ] = 264}$

$\implies \sf{4n^2 - 20n - 264 = 0}$

$\implies \sf{n^2 - 5n - 66 = 0}$

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Solving the Equation using Middle Term Splitting Method

We need to find two numbers such that their product is 66 and difference is 5

Two such numbers are 11 and 6

$\implies \sf{n^2 - 11n + 6n - 66 = 0}$

$\implies \sf{n \: (n - 11) + 6 \: (n - 11) = 0}$

$\implies \sf{(n-11) \: (n+6) = 0}$

$\sf{ }$

Either ( n - 11 ) = 0 $\implies$ n = 11

Or ( n + 6 ) = 0 $\implies$ n = ( - 6 )

But number of terms cannot be negative Hence n = - 6 is rejected

Hence $\boxed{\sf{n = 11}}$

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$\huge\underline{\sf{\red{A}\orange{n}\green{s}\pink{w}\blue{e}\purple{r}}}$

$\boxed{\sf{Number \: of \: terms \: with \: sum 132 = 11 \: terms }}$

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