Q23 / 118 · NTA Abhyas (JEE Main) 2020 · Mathematics · Sequences and Series

If a1+a5+a10+a15+a20+a24=225, then the sum of the first 24 terms of the arithmetic progression a1, a2, a3..... is equal to

  • A
    450
  • B
    675
  • Correct answer
    C
    900
  • D
    1200

Solution

S=a1+a2+a3+....+a23+a24

S=242a1+a24=12a1+a24

Given that, a1+a5+a10+a15+a20+a24=225

⇒a1+a1+4d+a1+9d+a24-9d+a24-4d+a24=225

⇒3a1+a24=225

⇒a1+a24=75

⇒S=12a1+a24=12×75=900

Hence, the correct answer is option: 'C'.

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