Question

The number of terms common between the series 1 2 4 8 to 100 terms and 1 4 7 10 to 100 terms is

Answer 1

Note: Number of common terms will be equal to number of terms with [LCM of two common differences] as common difference

Answer 2

ANSWER:

The number of terms common between the series 1 2 4 8 to 100 terms and 1 4 7 10 to 100 terms is 5

SOLUTION:

Given, two series are

1 2 4 8 ……… 100

1 4 7 10 …….. 100

We have to find the number of terms which are common in both the given series.

As we can see that, first series is in geometric progression, with common ratio 2.

Then, nth term of the series is $\mathrm{t}_{\mathrm{n}}=\mathrm{a} \cdot \mathrm{r}^{\mathrm{n}-1}$ where a is first term and r is common ratio.

And, the second series is in arithmetic progression, with common difference 3.

Then, “m” th term of the series is $\mathrm{t}_{\mathrm{m}}=\mathrm{a}+(\mathrm{m}-1) \mathrm{d}$ where a is first term and d is common difference.

Now, equate the nth term of G.P and mth term of A.P

$\mathrm{t}_{n}=\mathrm{t}_{\mathrm{m}}$

$a \mathrm{r}^{n-1}=a+(m-1) d$

$\begin{array}{l}{1.2^{n-1}=1+(m-1)^3} \\ {2^{n-1}=1+3 m-3} \\ {2^{n-1}=3 m-2} \\ {3 m=2^{n-1}+2}\end{array}$

Here, value of $2^{n-1}+2$ should be multiple of 3 as "m" can not be a fractional value.

So now, the set of values of $2^{n-1}+2$ are 3, 6, 9, 12, 15, 18, 21, ..300 [as we are considering 100 terms]

Now, the set of values of  $2^{n-1}$ are 3 – 2, 6 – 2, 9 – 2, ….. 300 – 2.

Then, set of values of $2^{n-1}$ are 1, 4, 7, 10, …. 298.

We know that, values of n are natural numbers, so we have to consider only the terms which can be written as 2 power something.

$\begin{array}{l}{\text { Then, value set of } 2^{n-1} \text { is } 1,4,16,64,256 \text { . }} \\ {\text { Now, values set of } 2^{n-1} \text { is } 2^{0}, 2^{2}, 2^{4}, 2^{6}, 2^{8}}\end{array}$

So, values set of n – 1 = 0, 2, 4, 6, 8 → n = 1, 3, 5, 7, 9

Hence, there are 5 terms which are common in both the given series.