MCQ
If there are 15 terms in an A.P. whose first term is $\sqrt{2}$ and common difference is $2 \sqrt{2}$, then the last term is:
- A$31 \sqrt{2}$
- B$30 \sqrt{2}$
- ✓$29 \sqrt{2}$
- D$28 \sqrt{2}$
✓
Answer
Correct option: C.
$29 \sqrt{2}$
(c) $29 \sqrt{2}$
Explanation:
Here, $n=15, a=\sqrt{2}, d=2 \sqrt{2}$
$\therefore$ Last term $=t_{15}=a+14 d$
$\begin{array}{l}=\sqrt{2}+14 \times 2 \sqrt{2} \\ =29 \sqrt{2} .\end{array}$
Explanation:
Here, $n=15, a=\sqrt{2}, d=2 \sqrt{2}$
$\therefore$ Last term $=t_{15}=a+14 d$
$\begin{array}{l}=\sqrt{2}+14 \times 2 \sqrt{2} \\ =29 \sqrt{2} .\end{array}$
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