the sum of 3rd and 11th terms of an arithmetic progression is 34 .find the sum of 13 term of the series
Answers
The sum of first 13 terms = 221
Step-by-step explanation:
Let the first term be "a" and the common difference between the terms of that AS be "d".
From the properties of Arithmetic set
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. Thus 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
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