# Find the first term of an a.p. if the sum of the first 10 terms is 150 and d =3.

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Le a1, a2, ..., a10 are terms of an A.P

given S10 = a1+a2+...+10 = 150

But we know that:

s10 = (a1+a10)*10/2 = 150

==> a1+a10 = 150/5

==> a1+a10 = 30

==> a10 = 30 - a1......(1)

But a10 = a1 + 9*3 = a1 + 27

==> a10 = 27 + a1........(2)

from (1) and (2):

==> 30 - a1 = 27 + a1

==> 2a1 = 3

**==> a1= 3/2 **

The determine the first term a1 of an AP whose number of terms n = 10 and whose sum , Sn = 150 and common difference d = 3.

Sn = (2a1+(n-1)d)n/2

150 = (2a1+(10-1)3)10/2

Multiply both sides by 5:

30 = 2a1 + 9*3

Subtract 27:

30-27 = 2a1

a1 = 3/2.

Therefore the first term of the given AP = 3/2.

We'll calculate the sum of the first 10 terms of an arithmetic series using the formula:

S10 = (a1+a10)*10/2, where a1 is the first term and a10 is the 10th term of the sum.

We'll write the 10th term with respect to a1 and the common difference d.

a10 = a1 + (10-1)*d

a10 = a1 + 9*3

a10 = a1 + 27

We'll substitute a10 and the value of S10 into the formula of the sum:

150 = (a1 + a1 + 27)*5

150 = (2a1+27)*5

We'll remove the brackets and we'll get:

150 = 10a1 + 135

We'll subtract both sides 135:

10a1 = 150 - 135

10a1 = 15

We'll divide by 10:

**a1 = 1.5**

**The first term of the arithemtic progression, whose sum is 150 and common difference is d = 3, is a1 = 1.5.**