Sequences and Series 2 Question 5
6. $n _1, n _2, \ldots, n _p$ are $p$ positive integers, whose sum is an even number, then the number of odd integers among them is odd.
(1985, 1M)
Integer Answer Type Question
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Answer:
Correct Answer: 6. (a)
Solution:
- Clearly, the two digit number which leaves remainder 2 when divided by 7 is of the form $N=7 k+2$ [by Division Algorithm]
For,
$$ \begin{gathered} k=2, N=16 \\ k=3, N=23 \\ \vdots \\ k=13, N=93 \end{gathered} $$
$\therefore 12$ such numbers are possible and these numbers forms an AP.
Now,
$$ \begin{aligned} S=\frac{12}{2}[16+93]= & 654 \\ \because S _n= & \frac{n}{2}(a+l) \end{aligned} $$
Similarly, the two digit number which leaves remainder 5 when divided by 7 is of the form $N=7 k+5$
For $k=1, N=12$
$$ \begin{aligned} & k=2, N=19 \\ & \vdots \\ & k=13, N=96 \end{aligned} $$
$\therefore 13$ such numbers are possible and these numbers also forms an AP.
Now,
$$ \begin{aligned} S^{\prime}=\frac{13}{2}[12+96] & =702 \\ & \because S _n=\frac{n}{2}(a+l) \end{aligned} $$
Total sum $=S+S^{\prime}=654+702=1356$
