find the sum of 51terms of an AP whose second and third terms are 14 and 18res
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Answered by
0
Answer:
if 2nd term is 14
and 3 term is 18
The common difference is 3rd term minus 2nd term (the two are concecutive)
ie common difference is 4
if common difference is 4
then second term can be written as
a+d ie a+4=14
the a is 10
the sum of number is
51/2(2•10+(50-1) 4)
=706
Answered by
5
Answer:5610
Step-by-step explanation:
Since the 2nd term of AP is 14
a+(n-1)d=14
Where
a+(2-1)d=14
a+d=14----(1)
And the 3rd term of the AP is 18
a+(n-1)d=18
a+(3-1)d=18
a+2d=18----(2)
By subtracting (1) from (2)
We get
d=4, a=10
Formula for finding the sum of a specific nth term is
Sn=(n/2)(2a+(n-1)d)
In this case
S51=(51/2)(20+(50*4))
=(51/2)(20+200)
=(51/2)(220)
=5610
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