Sequences & Series
Sequences and Series
nta_pyq_2025_jan
Grade 11
Question:
Let $a_{1},a_{2},a_{3},\dots$ be a G.P.\ of increasing positive terms. If $a_{1}a_{5}=28$ and $a_{2}+a_{4}=29$, then $a_{6}$ is equal to:
Step-by-Step Solution
Key Concept: With $a_{1}=a, a_{n}=ar^{n-1}$: square the second relation and divide by the first to eliminate $a$; you get a quadratic in $r+1/r$ (or directly in $r$).
$a_{1}a_{5}=a\cdot ar^{4}=a^{2}r^{4}=28.$
$a_{2}+a_{4}=ar+ar^{3}=ar(1+r^{2})=29.$
Square the second and divide by the first:
$$\frac{a^{2}r^{2}(1+r^{2})^{2}}{a^{2}r^{4}}=\frac{(1+r^{2})^{2}}{r^{2}}=\frac{29^{2}}{28}.$$
So $\dfrac{1+r^{2}}{r}=\dfrac{29}{\sqrt{28}}$, giving $\sqrt{28}\,r^{2}-29r+\sqrt{28}=0$, whose roots are $r=\sqrt{28}$ or $r=\dfrac{1}{\sqrt{28}}.$ For an increasing sequence, $r=\sqrt{28}.$
Then $a^{2}=28/r^{4}=28/28^{2}=1/28\Rightarrow a=1/\sqrt{28}.$
$$a_{6}=ar^{5}=\frac{1}{\sqrt{28}}\cdot(\sqrt{28})^{5}=(\sqrt{28})^{4}=28^{2}=784.$$
Correct Answer: 4