Question

Here's an easy question for you!


Consider two AP'S :-

A1 : 2, 7, 12, 17, . . . , upto 500 terms.

A2 : 1, 8, 15, 22, . . . , upto 300 terms.

Find the number of common terms and last common term between these two A.P.​

Answer 1

$\large\underline{\sf{Solution-}}$

Consider two AP'S :-

A1 : 2, 7, 12, 17, . . . , upto 500 terms.

A2 : 1, 8, 15, 22, . . . , upto 300 terms.

Now, Consider first series

$\rm :\longmapsto\:2,7,12,17,22, - - - 500 \: terms$

Its an AP series with

• First term, a = 2

• Common difference, d = 7.

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

↝ nᵗʰ term of an arithmetic sequence is,

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{a_n\:=\:a\:+\:(n\:-\:1)\:d}}}}}} \\ \end{gathered}$

Wʜᴇʀᴇ,

• aₙ is the nᵗʰ term.

• a is the first term of the sequence.

• n is the no. of terms.

• d is the common difference.

Tʜᴜs,

$\rm :\longmapsto\:a_{500} = a + (500 - 1)d$

$\rm :\longmapsto\:a_{500} = 2 + (500 - 1)5$

$\rm :\longmapsto\:a_{500} = 2 + (499)5$

$\rm :\longmapsto\:a_{500} = 2 + 2495$

$\rm :\longmapsto\:a_{500} = 2497$

So, series is

$\rm :\longmapsto\:\boxed{ \tt{ \: 2,7,12,17,22, - - - ,2497 \: }}$

Now, Consider

$\rm :\longmapsto\:1,8,15,22, - - - 300 \: terms$

Its an AP series with

• First term, a = 1

• Common difference, d = 7

So,

$\rm :\longmapsto\:a_{300} = 1 + (300 - 1)7$

$\rm :\longmapsto\:a_{300} = 1 + (299)7$

$\rm :\longmapsto\:a_{300} = 1 + 2093$

$\rm :\longmapsto\:a_{300} = 2094$

So, series is

$\rm :\longmapsto\:\boxed{ \tt{ \: 1,8,15,22, - - -,2094 \: }}$

Now,

Common difference of first AP series = 5

Common difference of second AP series = 7

So, Common difference of Common terms = 5 × 7 = 35

and first term = 22

So, series become

$\rm :\longmapsto\:22,57,92, - - -$

So, to find number of terms, we assume that last term of the series is less than or equals to 2094.

So,

$\rm :\longmapsto\:22 + (n - 1)35 \leqslant 2094$

$\rm :\longmapsto\:(n - 1)35 \leqslant 2094 - 22$

$\rm :\longmapsto\:(n - 1)35 \leqslant 2072$

$\rm :\longmapsto\:n - 1 \leqslant 59.2$

$\rm :\longmapsto\:n \leqslant 59.2 + 1$

$\rm :\longmapsto\:n \leqslant 60.2$

$\bf\implies \:n = 60$

Thus,

$\rm \implies\:a_{60} = a + (60 - 1)d$

$\rm \implies\:a_{60} = a + 59d$

$\rm \implies\:a_{60} = 22 + 59 \times 35$

$\rm \implies\:a_{60} = 22 + 2065$

$\rm \implies\:a_{60} = 2087$

Hence,

$\begin{gathered}\begin{gathered}\bf\: \rm \implies\:\begin{cases} &\sf{n \: = \: 60} \\ \\ &\sf{a_{60} = 2087} \end{cases}\end{gathered}\end{gathered}$