Question

Let \( S_{n} \) denote the sum of first \( n \) terms of an arithmetic progression. If \( S_{20}=790 \) and \( S_{10}=145 \), then \( S_{15}-S_{5} \) is :

Answer 1

For n=20:

S₂₀=(20/2)(2A+19D)=790
Simplifying this:

10(2A+19D)=790⟹2A+19D=79

(Equation 1)

For n=10:
S₁₀=(10/2)(2A+9D)=145
Simplifying this:
5(2A+9D)=145⟹2A+9D=29

(Equation 2)

Solving the equations:
Now we have two equations:
 2A+19D=79 (Equation 1)
 2A+9D=29 (Equation 2)

We can subtract Equation 2 from Equation 1:
(2A+19D)−(2A+9D)=79−29
This simplifies to:
10D=50⟹D=5

Substitute D=5 back into Equation 2:
2A+9(5)=29⟹2A+45=29⟹2A=−16⟹A=−8

 Calculating S₁₅ and S₅:
Now we can find S₁₅ and S₅ :
- For S₁₅:
S₁₅=(15/2)(2A+14D)=(15/2)(2(−8)+14(5))
=15×27=405

- For S₅:
S₅=(5/2)(2A+4D)=(5/2)(2(−8)+4(5))
=5×2=10

6. Finding S₁₅−S₅:
Now, we can calculate:
S₁₅−S₅=405−10=395

The final answer is:
395

Answer 2

The sum of the first n terms of an arithmetic progression (A.P.) is given by:

\(S_n = \frac{n}{2} (2A + (n - 1)D)\)

where A is the first term and D is the common difference.

For n = 20:

\(S_{20} = \frac{20}{2} (2A + 19D) = 790\)

10(2A + 19D) = 790

2A + 19D = 79 ............(i)

For n = 10:

\(S_{10} = \frac{10}{2} (2A + 9D) = 145\)

5(2A + 9D) = 145

2A + 9D = 29 ............(ii)

We can subtract Equation 2 from Equation 1:

(2A + 19D) − (2A + 9D) = 79 − 29

10D = 50

D = 5

Substitute D=5 back into Equation 2:

2A + 9(5) = 29

2A + 45 = 29

2A = 29 − 45

2A = −16

A = −8

For S₁₅:

\(S_{15} = \frac{15}{2}(2A + 14D)\)

\(= \frac{15}{2} (2(-8) + 14(5))\)

\(= \frac{15}{2} (-16 + 70)\)

\(\frac{15}{2} \times 54 = 15 \times 24 = 405\)

For S₅:

\(S_5 = \frac{5}{2}(2A + 4D)\)

\(= \frac{5}{2} (2(-8) + 4(5))\)

\(= \frac{5}{2} (-16 + 20) \)

\(= \frac{5}{2} \times 4 = 5 \times 2 = 10\)

S₁₅ − S₅

= 405 − 10

= 395