Suppose a1, a2, a3, a4 be in an arithmetico-geometric progression. If the common ratio of the corresponding geometric pr

Question

Suppose a1, a2, a3, a4 be in an arithmetico-geometric progression. If the common ratio of the corresponding geometric progression in 2 and the sum of all 5 terms of the arithmetico-geometric…

Accepted Answer

Given: The terms are in an arithmetico-geometric progression and the common ratio of the corresponding geometric progression is $2$ . The sum of all $5$ terms is $\frac{49}{2}$ .

Find: $a_4$ .

From the given solution, the five terms are

\frac{a-2d}{4}, \frac{a-d}{2}, a, 2(a+d), 4(a+2d)

Given that $a = 2$ , substitute into these terms:

\frac{a-2d}{4}, \frac{a-d}{2}, a, 2(a+d), 4(a+2d) \Rightarrow \frac{2-2d}{4}, \frac{2-d}{2}, 2, 2(2+d), 4(2+2d)

Now use the sum of the $5$ terms:

\left(\frac{1}{4}+\frac{1}{2}+1+6\right) \times 2 + (-1+2+8)d = \frac{49}{2}

2\left(\frac{3}{4}+7\right) + 9d = \frac{49}{2}

2 \times \frac{31}{4} + 9d = \frac{49}{2}

\frac{62}{4} + 9d = \frac{49}{2}

Now solve for $d$ :

9d = \frac{49}{2} - \frac{62}{4} = \frac{98}{4} - \frac{62}{4} = \frac{36}{4} = 9

Thus,

d = 1

Now substitute $d = 1$ into

a_4 = 4(a+2d)

So,

a_4 = 4(2+2 \times 1) = 4(2+2) = 4 \times 4 = 16

Therefore, the value of $a_4$ is $16$ .

Answer

Step-by-step solution

Common mistakes

Practice more Geometric Progression (GP) questions

Related questions