Question

Find the A.P. whose 7th and 13th terms are respectively $34$ and $64$. If $a,b,c$ are in A.P, prove that $b+c,c+a,a+b$ are also in A.P. The sum of four integers in A.P is 24 and their product is $945$. Find the numbers.

Answer 1

$T_n=a+(n-1)d$

$T_7=a+6d=34$....(1)

$T_{13}=a+12d=64$...(2)

solving (2) and (1) gives, $d=5,a=4$

Therefore the series is: $4,9,14,19,......$

---------------------------------------------------------------------------------

Since a,b,c are in A.P.

$2b=a+c$

we have to prove that $2(a+c)=(b+c)+(a+b)$

$2a+2c=b+c+a+b$

$a+c=2b$

Therefore (a+b),(b+c),(c+a) are in A.P.

----------------------------------------------------------------------------------

Let the numbers be $(a-3d),(a-d),(a+d),(a+3d)$

Given,

$a-3d+a-d+a+d+a+3d=24$

$4a=24\Rightarrow a=6$

$(a-3d)(a-d)(a+d)(a+3d)=945$

$(6-3d)(6-d)(6+d)(6+3d)=945$

$(36-9d^2)(36-d^2)=945$

$9d^4-360d^2+9d^4-945=0$

$(d^2-1)(d^2-39)=0$

$d=1,\sqrt{39}$

$\therefore d=1$

The required numbers are $(6-3(1\left)\right),(6-1),(6+1),(6+3(1\left)\right)=3,5,7,9$