Question

Let an odd number of terms of an A.P. be such that the sum and the product of its first and last terms are 10 and 0 respectively. If the common difference is 1/10, then the number of terms n is (a) 99, (b) 101, (c) 103, (d) 191​

Answer 1

Answer:

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Answer 2

Given:

Sum of first and last terms of AP = 10

Product of first and last terms of AP = 0

Common difference = $\frac{1}{10}$

To find:

Number of terms.

Solution:

In an AP, let the first term be $a$ and the last term be $a_{n}$. Then, the difference between two consecutive terms called common difference is $d$.

$a_{n}=a+(n-1)d...(1)$

where $n$ is the number of terms.

Since, the problem says that there odd number of terms in the AP, that means $n$ is odd. Let $2n-1$ be the total number of terms in the AP.

$a+a_{n}=10...(2)$

$a*a_{n}=0$

If the product of first and last terms is 0, then one of the terms is 0.

Let  $a$ be 0.

Substituting this value in equation (2)

$a+a_{n}=10$

$0+a_{n}=10$

$a_{n}=10$

Substituting the values of $a,a_{n},d$ in equation (1)

$a_{n}=a+(n-1)d$

$10=0+(n-1)\frac{1}{10}$

$\\ 10=\frac{n-1}{10}$

On cross-multiplying,

$100=n-1$

$100+1=n=101$

Here, the no. of terms we obtained is 101 which is an odd number.

∴ option (b) 101 is the correct answer.

The no. of terms in an AP whose sum and product of its first and last terms are 10 and 0 respectively with common difference $\frac{1}{10}$ is 101.

The correct option is (b) 101.