Question

An arithmetic progression consists of 30 terms. The sum of its first 10 terms 185. 21 term is 15 more than the 16 term. Find the sum of the terms of the arithmetic progression.

Answer 1

Identify the formula for the sum of the first n terms

The sum of the first n terms (S_n) of an arithmetic progression (AP) can be calculated using the formula:

\(S_n=\frac{n}{2}\times(2 a+(n-1) d)\)

where:

n= number of terms

a = first term

d = common difference

Use the information given for the first 10 terms

We know that the sum of the first 10 terms (s₁₀) is 185. 

Therefore:

\(S_{10}=\frac{10}{2} \times(2 a+(10-1) d)=185\)

This simplifies to:

\(5(2 a+9 d)=185\)

Dividing both sides by 5 gives:

\(2 a+9 d=37\)

(Equation 1)

Use the information about the 21st and 16th terms

The 21st term (T₂₁) is given by:

\(T_{21}=a+20 d\)

The 16th term (T₁₆) is given by:

\(T_{16}=a+15 d\)

According to the problem, T₂₁ is 15 more than T₁₆ :

\(a+20 d=(a+15 d)+15\)

Simplifying this gives:

\(a+20 d=a+15 d+15\)

Subtracting a and 15d from both sides results in:

\(5 d=15\)

Thus, we find:

\(d = 3\)

(Equation 2)

Substitute d back into Equation 1

Now we substitute d = 3 into Equation 1:

\(2 a+9(3)=37\)

This simplifies to:

\(2 a+27=37\)

Subtracting 27 from both sides gives:

\(2 a=10\)

Thus:

\(a=5\)

(Equation 3)

Calculate the sum of all 30 terms

Now that we have both a and d, we can find the sum of all 30 terms (S₃₀):

\(S_{30}=\frac{30}{2} \times(2 a+(30-1) d)\)

Substituting a = 5 and d = 3 :

\(S_{30}=15\times(2(5)+29(3))\)

Calculating inside the parentheses:

\(S_{30}=15 \times(10+87)=15 \times 97=1455\)

The sum of the terms of the arithmetic progression is 1455.