Question

ICBSE
In an AP. If the first term is 22. the common difference is -4 and the sum to n terms is 64,find n.​

Answer 1

GIVEN :–

• First term of A.P.(a) = 22

• Common difference(d)= -4

• Sum of n terms = 64

TO FIND :–

• Value of 'n' = ?

SOLUTION :–

• We know that Sum of n terms –

$\\ \: \dashrightarrow \: \large{ \boxed{\bf S_n = \dfrac{n}{2}[2a + (n - 1)d]}}$

• Put the values –

$\\\implies\bf 64 = \dfrac{n}{2}[2(22)+ (n - 1)( - 4)]$

$\\\implies\bf 64 = \dfrac{2n}{2}[22+ (n - 1)( - 2)]$

$\\\implies\bf 64 =n(22 - 2n + 2)$

$\\\implies\bf 64 =n(24 - 2n)$

$\\\implies\bf 64 =24n - 2n^{2}$

$\\\implies\bf 2n^{2} - 24n + 64 =0$

$\\\implies\bf n^{2} - 12n +32 =0$

$\\\implies\bf n^{2} - 8n - 4n +32 =0$

$\\\implies\bf n(n - 8) - 4(n - 8)=0$

$\\\implies\bf (n - 4)(n - 8)=0$

$\\\implies\bf n = 4,8$

• Hence –

$\\\implies \large{ \boxed{\bf n = 4 \: \: or \: \: 8}}$

Answer 2

$\huge{\boxed{\rm{Question}}}$

In an AP if the first term is 22. The common difference is -4 and the sum to n terms is 64. Find the value of n.

$\huge{\boxed{\rm{Answer}}}$

$\large{\boxed{\boxed{\sf{Given \: that}}}}$

• The 1st term lf an AP is 22.

• Common difference between them = -4

• The sum of n terms = 64

$\large{\boxed{\boxed{\sf{To \: find}}}}$

• Value of n.

$\large{\boxed{\boxed{\sf{Solution}}}}$

• Value of n = 4 , 8

$\large{\boxed{\boxed{\sf{Using \: formula}}}}$

• To find sum of n terms

$S_{n}$ = $\frac{n}{2}$ [2a+ (n-a)d]

$\large{\boxed{\boxed{\sf{What \: does \: the \: question \: says}}}}$

$\large{\boxed{\boxed{\sf{Let's \: understand \: the \: concept \: 1st}}}}$

• This question says that in an AP the first term is 22. After that it say that the common difference is -4 and the sum to n terms is 64 Afterwards it ask us to find the value of n.

$\large{\boxed{\boxed{\sf{How \: to \: solve \: this \: question}}}}$

$\large{\boxed{\boxed{\sf{Let's \: see \: the \: procedure \: now}}}}$

• To solve this problem, we have to use the formula afterthat putting the values according to this formula. We get our final result that is n = 4 , 8

$\large{\boxed{\boxed{\sf{Full \: solution}}}}$

$S_{n}$ = $\frac{n}{2}$ [2a+ (n-a)d]

Putting the values we get the following results,

64 = $\frac{n}{2}$ [ 22 + (n-1)(-2) ]

• 2 and 2 cancel each other. After adding them we get,

64 = n(22 -2n + 2)

64 = n(22 + 2 -2n)

64 = n(24 - 2n)

64 = 24n - 2n²

• + = -

2n² - 24n + 64 = 0

• Divide 24 by 2 we get 12.
• Divide 64 by 2 we get 32.

n² - 12n + 32 = 0

n² - 8n - 4n + 32 = 0

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

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

(n4) (n8) = 0

n = 4 , 8

Therefore, n = 4 , 8

$\bold{\pink{\fbox{\pink{n = 4 , 8}}}}$