The first term of an A.P. is – 5 and the 6th term is 45. Find S6
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First term a=-5, let say total no. of terms =n,
b is the common difference.
Now we can write last term as ‘a+(n-1)b=45’
Sum of n terms as ‘na+(n(n-1)b)/2=120’
By solving thease two equation n=6, b=10.
-5+(n-1)b=45
(n-1)b=50
n*(-5)+(n*50)/2=120
n*(-5)+n*25=120
n=120/20=6
From (n-1)b=50
(6–1)b=50
b=10.
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Answer:
S6= 120
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