Question

If $a_{1},a_{2},a_{3},.....,a_{24}$ are in arithmetic progression and $a_{1} + a_{5} + a_{10} + a_{15} + a_{20} + a_{24} = 225$, then $a_{1} + a_{2} + a_{3} + ..... + a_{23} + a_{24} =$

• A. 909
• B. 75
• C. 750
• D. 900

Answer 1

Answer: (D)

900

Answer 2

$a_{1} + a_{5} + a_{10} + a_{15} + a_{20} + a_{24} = 225$

⇒ $(a_{1} + a_{24}) + (a_{5} + a_{20}) + (a_{10} + a_{15}) = 225$

⇒ $3(a_{1} + a_{24}) = 225$ ⇒ $a_{1} + a_{24} = 75$

(∵ In an A.P. the sum of the terms equidistant from the beginning and the end is same and is equal to the sum of first and last term)

$a_{1} + a_{2} + ..... + a_{24} = \frac{24}{2}(a_{1} + a_{24}) = 12 \times 75 = 900$