in an A.P. ,if a16=p,then find the sum of first 31 terms
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Answered by
71
Hi,
Let a , d are first term and common difference of an A.P
*********************************************************
nth term of A P = an = a + ( n - 1 ) d ----------( 1 )
*********************************************************
Given,
a 16 = p
a + 15 d = p ---------------( 2 ) from ( 1 )
*********************************************************************
Sum of first n terms an A P
S n = n/2 [ 2a + ( n - 1 ) d ]
*********************************************************************
Given ,
S 31 = 31 /2 [ 2 a + ( 31 - 1 ) d ]
S 31 = 31 / 2 × [ 2 a + 30 d ]
= 31 / 2 × 2 [ a + 15 d ]
= 31 ( a + 15 d )
= 31 p [ from equation ( 2 )
Therefore ,
S 31 = 31 p
I hope this will useful to you.
*****
Let a , d are first term and common difference of an A.P
*********************************************************
nth term of A P = an = a + ( n - 1 ) d ----------( 1 )
*********************************************************
Given,
a 16 = p
a + 15 d = p ---------------( 2 ) from ( 1 )
*********************************************************************
Sum of first n terms an A P
S n = n/2 [ 2a + ( n - 1 ) d ]
*********************************************************************
Given ,
S 31 = 31 /2 [ 2 a + ( 31 - 1 ) d ]
S 31 = 31 / 2 × [ 2 a + 30 d ]
= 31 / 2 × 2 [ a + 15 d ]
= 31 ( a + 15 d )
= 31 p [ from equation ( 2 )
Therefore ,
S 31 = 31 p
I hope this will useful to you.
*****
Answered by
11
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