Question

find the sum of first 22 terms of the AP whose first term is 8 and common difference is -5​

Answer 1

Answer:

$\huge\tt{\boxed{\underbrace{\overbrace{\bold\color{blue}{☛AnSwEr☚}}}}}$

In the given problem, we need to find the number of terms of an A.P. Let us take the number of terms as n.

Here, we are given that,

38

a = 22

d = -4

S_n= 6

So, as we know the formula for the sum of n terms of an A.P. is given by,

So, as we know the formula for the sum of n terms of an A.P. is given by,

'S_n= n/2 [2a + (n - 1)d]

38

Where; a = first term for the given A.P.

d = common difference of the given A.P.

n = number of terms

So, using the formula we get,

A_n= n/2 [2(22) + (n - 1)(-4)]

38

64 = n/2[44 - 4 n + 4] =

64(2) = n(48 - 4 n) =

128 = 48n - 4n^2

Further rearranging the terms, we get a quadratic equation,

4n^2 - 48n + 128 = 0

On taking 4 common we get

'n^2 - 12n + 32 = 0

Further, on solving the equation for n by splitting the middle term, we get,

'n^2 - 12n + 32 = 0

`n^2 - 8n -4n + 32 = 0

n(n 8) - 4(n - 8) = 0

(n - 8)(n 4) = 0

38

So, we get

(n - 8) = 0

n = 8

Also

(n - 4) = 0

n = 4

$\huge\orange{\boxed{\blue{\bold{\fcolorbox{orange}{yellow}{\orange{Hope\:It\:Helps༒}}}}}}$$\huge\orange{\underline\blue{\underline\green{\underline{ \mathtt\blue{Please}\:\mathtt\green{Mark}\:\mathtt\orange{As}}}}}$

$\huge\orange{\underline\blue{\underline\green{\underline{ \mathtt\red{Brilliant}\:\mathtt\purple{Answers}}}}}$

$\huge\orange{\underline\blue{\underline\green{\underline{ \mathcal\orange{Thank}\:\mathcal\green{You}}}}₪}$

Answer 2

In the given problem, we need to find the number of terms of an A.P. Let us take the number of terms as n.

Here, we are given that,

a = 22

d = -4

S_n= 6

So, as we know the formula for the sum of n terms of an A.P. is given by,

`S_n = n/2 [2a + (n - 1)d]`

Where; a = first term for the given A.P.

d = common difference of the given A.P.

n = number of terms

So, using the formula we get,

`S_n= n/2 [2(22) + (n - 1)(-4)]`

`64 = n/2[44 - 4n + 4]`

64(2) = n(48 - 4n)

`128 = 48n - 4n^2`

Further rearranging the terms, we get a quadratic equation,

`4n^2 - 48n + 128 = 0`

On taking 4 common we get

`n^2 - 12n + 32 = 0`

Further, on solving the equation for n by splitting the middle term, we get,

`n^2 - 12n + 32 = 0`

`n^2 - 8n -4n + 32 = 0`

n(n - 8) - 4(n - 8) = 0

(n - 8)(n - 4) = 0

So, we get,

(n - 8) = 0

n = 8

Also

(n - 4) = 0

n = 4

Therefore n = 4 or 8