Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# The sum of four consecutive terms which are in an arithmetic progression is 32 and the ratio of the product of the first and the last term to the product of two middle terms is 7:15. Find the number.

Last updated date: 20th Jun 2024
Total views: 441k
Views today: 11.41k
Verified
441k+ views
Hint: To solve the question, using the given information of the sum of the terms and ratio of the product of the first and the last term to the product of two middle terms form two equations and solve the equations to find the unknown terms of the arithmetic progression.

Let $a-3d,a-d,a+d,a+3d$ be the four consecutive terms which are in arithmetic progression (AP) where $a,d$ are integers.
The given value of the sum of these four consecutive terms of AP is equal to 32.
The sum of the four consecutive terms of AP $=a-3d+a-d+a+d+a+3d=4a$.
$\Rightarrow 4a=32$
$a=\dfrac{32}{4}=8$
The given ratio of product of the first and the last term to the product of two middle terms is equal to 7:15.
The calculated ratio of product of the first and the last term to the product of two middle terms
$=\dfrac{\left( a-3d \right)\left( a+3d \right)}{\left( a-d \right)\left( a+d \right)}$
We know the formula $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$
$=\dfrac{\left( {{a}^{2}}-{{\left( 3d \right)}^{2}} \right)}{\left( {{a}^{2}}-{{d}^{2}} \right)}$
$=\dfrac{\left( {{a}^{2}}-9{{d}^{2}} \right)}{\left( {{a}^{2}}-{{d}^{2}} \right)}$
$\Rightarrow \dfrac{7}{15}=\dfrac{{{a}^{2}}-9{{d}^{2}}}{{{a}^{2}}-{{d}^{2}}}$
By cross multiplication we get,
$7\left( {{a}^{2}}-{{d}^{2}} \right)=15\left( {{a}^{2}}-9{{d}^{2}} \right)$
$135{{d}^{2}}-7{{d}^{2}}=15{{a}^{2}}-7{{a}^{2}}$
$128{{d}^{2}}=8{{a}^{2}}$
$16{{d}^{2}}={{a}^{2}}$
${{d}^{2}}=\dfrac{{{a}^{2}}}{16}={{\left( \dfrac{a}{4} \right)}^{2}}$
$\Rightarrow d=\pm \dfrac{a}{4}$
By substituting the value $a$ in the above equation, we get
$d=\pm \dfrac{8}{4}=\pm 2$
Thus, the values of the four consecutive terms of arithmetic progression are as follows
For d = 2
$a-3d=\left( 8-3\times 2 \right)=8-6=2$
$a-d=\left( 8-2 \right)=6$
$a+d=\left( 8+2 \right)=10$
$a+3d=\left( 8+3\times 2 \right)=8+6=14$
For d = -2
$a-3d=\left( 8-3\times (-2) \right)=8+6=14$
$a-d=\left( 8-(-2) \right)=8+2=10$
$a+d=\left( 8+(-2) \right)=8-2=6$
$a+3d=\left( 8+3\times (-2) \right)=8-6=2$
Thus, the four consecutive terms which are in an arithmetic progression are 2,6,10,14 or 14,10,6,2.

Note: The possibility of mistake can be the calculations as the procedure of solving involves multiple steps to solve. The other mistake can be not using both the calculated positive and negative values of d , since a and d are integers.