Question

If the first term of an A.P. is \(2\) and the sum of first five terms is equal to one fourth of the sum of the next five terms, then the sum of the first \(30\) terms is

• A. \(2550\)
• B. \(3000\)
• C. \(-2550\)
• D. \(-3000\)

Hint:

For arithmetic progressions:
Use \(S_n=\dfrac{n}{2}[2a+(n-1)d]\)
Sum of consecutive blocks can be found by subtraction
Always substitute the common difference carefully

Answer 1

The Correct Option is C

Step 1: Given first term \(a=2\). Let common difference be \(d\). Sum of first \(n\) terms of an A.P.: \[ S_n=\frac{n}{2}\,[2a+(n-1)d] \]
Step 2: Sum of first five terms: \[ S_5=\frac{5}{2}[2(2)+4d]=\frac{5}{2}(4+4d)=10+10d \]
Step 3: Sum of first ten terms: \[ S_{10}=\frac{10}{2}[2(2)+9d]=5(4+9d)=20+45d \] Sum of next five terms: \[ S_{6\text{ to }10}=S_{10}-S_5=(20+45d)-(10+10d)=10+35d \]
Step 4: Given condition: \[ S_5=\frac{1}{4}S_{6\text{ to }10} \] \[ 10+10d=\frac{1}{4}(10+35d) \] \[ 40+40d=10+35d \] \[ 5d=-30 \Rightarrow d=-6 \]
Step 5: Now find the sum of first \(30\) terms: \[ S_{30}=\frac{30}{2}[2(2)+29(-6)] \] \[ S_{30}=15(4-174)=15(-170)=-2550 \]