Question

if the sum of n terms of an arithemetic prograssion 2,5,8,.... is equal to the sum of first n terms of another arithemetic prograssion 57,59,61,... then find the value of n?​........?

Answer 1

Given:- Sum of nth terms of AP 2,5,8 is equal to sum of nth terms of AP 57,59,61.

To Find:- value of n

Solution :-

⇒Sum of n terms of AP 2,5,8 will be,

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

= n/2[ 2(2) + (n-1)3]

= n/2 [ 4 + 3n -3 ]

= n/2 [ 3n + 1 ] ....1

⇒Sum of n terms of AP 57,59,61 will be,

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

= n/2[ 2(57) + (n-1)2]

= n/2 [ 114 + 2n -2 ]

= n/2 [ 2n + 112 ] ....2

⇒Now, equation 1 and 2 are equal,

n/2 [ 3n + 1 ] = n/2 [ 2n + 112 ]

3n + 1 = 2n + 112

n = 111

Therefore, value of n is 111.

Answer 2

Hello,Buddy!!

Refer The Attachment ⤴️

• The Value of n ➪ $\bold{11}$

Formulas Used:-

• $\sf{S_{n} = \frac{n}{2}[2a+(n-1)d]}$

$\boxed{\tt{@MrSovereign♡}}$

Hope This Helps!!