Question

the sum of 3rd and 11th terms of an arithmetic progression is 34 .find the sum of 13 term of the series​

Answer 1

Answer:

Required sum of first 13 terms is 221.

Step-by-step-explanation:

Let the first term of that AS be a and the common difference between the terms of that AS be d.

From the properties of AS

nth term = a + ( n - 1 )d, where a is the first term, n is the number of terms and d is the common difference between the terms.

Sum of n terms = ( n / 2 )[ 2a + ( n - 1 )d ].

Here,

Sum of 3rd term and 11th term is 34.

= > 3rd term + 11th term = 34

= > [ a + ( 3 - 1 )d ] + [ a + ( 11 - 1 )d ] = 34

= > a + 2d + a + 10d = 34

= > 2a + 12d = 34

We have to find the sum of first 13 terms :

= > Sum of first 13 terms = ( 13 / 2 )[ 2a + ( 13 : 1 )d ]

= > Sum of first 13 terms = ( 13 / 2 )[ 2a + 12d ]

= > Sum of first 13 terms = ( 13 / 2 ) x ( 34 ) { from above }

= > Sum of first 13 terms = 221

Hence the required sum of first 13 terms is 221.

Answer 2

I hope it helps u guys...