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Model Paper 1 — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsModel Paper 11 Mark
MCQ
$(3+6+12+\ldots+1536)=?$
  • A
    1023
  • B
    3069
  • C
    2046
  • D
    4092
✓

Answer

(b) 3069 
Explanation: This is a GP in which $a =3, r =\frac{6}{3}=2$ and $l =1536$
$\therefore$ required sum $=\frac{(l r-a)}{(r-1)}=\frac{(1536 \times 2-3)}{(2-1)}=(3072-3)=3069$.

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