Question

Let $A.P.(a,d)$ denotes the set of all the terms of an infinite arithmetic progression with first term $a$ and common difference $d>0$ . If $A.P.(1,4)\cap A.P.(1,5)\cap A.P.(3,3)=A.P.(a,d)$ then

• A. $a=16$
• B. $d=60$
• C. sum of first $4$ terms of $A.P.(a,d)$ is $402$
• D. $a+d=81$

Answer 1

Answer: (B), (D)

The correct options are
B $d=60$
D $a+d=81$

Let $n^{th}$ of first $A.P.$, $m^{th}$ of second $A.P.$, $k^{th}$ of third $A.P.$ be equal, then

$1+(n-1)4=1+(m-1)5=3+(k-1)3$
$\Rightarrow 4n-3=5m-4=3k$
$\Rightarrow m=\frac{4n+1}{5}\Rightarrow k=\frac{4n-3}{3}$
Since, $n,m$ and $k$ are the number of terms, therefore $n,m,k\in N$
$\Rightarrow n=1,6,11,16,21,...$ for $m\in N$

Similarly,
$\Rightarrow n=3,6,9,12,...$ for $k\in N$

Least common value of $n$ is $6$
$1+(n-1)4=1+5\cdot 4\therefore a=21$

Common difference $d=L.C.M.(4,5,3)=60$
Sum of first four terms,
$=\frac{4}{2}(2\times 21+3\times 60)=444$