on 1st january 2016,sanika decides to save
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here,a=10
d=1
and since the days between 1 Jan. and 31 Dec. are 365
therefore, n=365
we have to find the total savings till the date
so,it means we have to find the S365
let's go,
S365=n÷2[2a+(n-1)d]
S365=365÷2[20+(364)1]
S365=365(10+182)=365×192
S365=70080
Therefore, on 31st December 2016 Sanika's total savings would be 70080 rupees.
PLZ MARK IT BRAINLIEST,I TRIED MY BEST!!!
HOPE IT HELPS!! HAVE A NICE DAY!!!
d=1
and since the days between 1 Jan. and 31 Dec. are 365
therefore, n=365
we have to find the total savings till the date
so,it means we have to find the S365
let's go,
S365=n÷2[2a+(n-1)d]
S365=365÷2[20+(364)1]
S365=365(10+182)=365×192
S365=70080
Therefore, on 31st December 2016 Sanika's total savings would be 70080 rupees.
PLZ MARK IT BRAINLIEST,I TRIED MY BEST!!!
HOPE IT HELPS!! HAVE A NICE DAY!!!
Priyanshu3008:
Mark it brainliest & tell me how was my answer,thanks!!!
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