Question

The first term of an AP is 3 the last term is 83 and the sum of all it's terms is 903.
Find the number of terms and the common difference.

Answer 1

here is your answer brother

a=3

an=83

Sn=903

Sn=n/2(a+an)

903=n/2(3+83)

903*2=86n

1806=86n

n=1806/86

n=21

now

Sn=n/2(2a+(n-1)d)

903=21/2(2*3+20d)

903*2/21=6+20d

1806/21=6+20d

86=6+20d

86-6=20d

80=20d

d=80/20

d=4

hope it will help full to you..

plzz mark as brainlist

Answer 2

GIVEN,

the first term of AP=3, the last term=83, the sum of all terms=903.

TO FIND,

the number of terms and the common difference in between.

SOLUTION,

Here we are given,

A=3, Aₙ=83, Sₙ=903

we know that the formula for the sum of 'n' terms is

Sₙ= $\frac{n}{2} (a+an)$

substituting the given values in the formula,

Sₙ= $\frac{n}{2}(3+83)$

903=$\frac{n}{2}$(86)

903*2= 86n

1806=86n

making 'n' the subject of the equation, we get

n= $\frac{1806}{86}$

∴ n= 21

now using,

Sₙ= $\frac{n}{2} [2a+(n-1)d]$

substituting the values to find 'd'

903=$\frac{21}{2}[2*3 + 20d]$

903 * $\frac{2}{21} = 6+20d$

simplifying,

$\frac{1806}{21}=6+20d\\\\86= 6+20d\\\\86-6= 20d\\\\d= \frac{80}{20} \\\\d=4$

HENCE NUMBER OF TERMS IS 21 AND THE COMMON DIFFERENCE IS 4.