Q.The 4th term of a G.P. is square of its second term, and the first term is $-3$. Determine its 7th term.

Summing the geometric series 2+4+8+...+2^10 (10 generations) gives 2046 ancestors.

The number of ancestors in generation k is $2^k$ (2 parents, 4 grandparents, 8 great-grandparents, ...). Over 10 generations, the total is the geometric series:

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Using the G.P. condition (middle term)² = product of the outer terms, x = 0 or x = 16.

For x+9, x−6, 4 to be in G.P., $(x-6)^2 = (x+9)(4)$.

Expand: $x^2-12x+36 = 4x+36$

$x^2-12x-4x = 0$

$x^2-16x = 0$

$x(x-16)=0$

So $x=0$ or $x=16$.

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For the G.P. $1, \tfrac13, \tfrac19,\ldots$ with $a=1$, $r=\tfrac13$, the infinite sum is $S_\infty = \dfrac{a}{1-r} = \dfrac{3}{2}$.

Here $a = 1$ and the common ratio $r = \dfrac{1/3}{1} = \dfrac13$. Since $|r| = \tfrac13 < 1$, the series converges and …

Applying $b^2=ac$ to $-\tfrac27, x, -\tfrac72$ gives $x^2=1$, i.e. $x=\pm1$, exactly matching the Assertion — so both statements are true and (R) explains (A).

For three numbers $a,b,c$ to be in G.P., the middle term squared must equal the product of the outer terms: $b^2 = ac$ — this is exactly Reason (R), which is a true, standard fact about G.P.s.

Apply it to $-\dfrac27,\ x,\ -\dfrac72$: treating $x$ as the middle term,

$x^2 = \left(-\dfrac27\right)\left(-\dfrac72\right) = 1 \implies x = \pm 1.$

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The value of $n$ that makes $(a^{n+1}+b^{n+1})/(a^n+b^n)$ equal to the geometric mean $\sqrt{ab}$ of $a$ and $b$ is $n=-\dfrac12$.

The G.M. between $a$ and $b$ is $\sqrt{ab}$. We are told

$\dfrac{a^{n+1}+b^{n+1}}{a^{n}+b^{n}} = \sqrt{ab}$

Step 1: Try $n=-\dfrac12$.

$\dfrac{a^{1/2}+b^{1/2}}{a^{-1/2}+b^{-1/2}}$

Multiply numerator and denominator by $\sqrt{ab}$:

$\dfrac{\sqrt{ab}(a^{1/2}+b^{1/2})}{\sqrt{ab}\,a^{-1/2} + \sqrt{ab}\,b^{-1/2}} = \dfrac{\sqrt{ab}(\sqrt a+\sqrt b)}{\sqrt b + \sqrt a}$

The factor $(\sqrt a+\sqrt b)$ cancels top and bottom, leaving exactly $\sqrt{ab}$ — confirming $n=-\dfrac12$ works. …

Using $x^2 = \left(-\dfrac27\right)\left(-\dfrac72\right)$ gives $x^2=1$, so $x=\pm1$.

Three numbers $a,b,c$ are in G.P. exactly when $b^2=ac$ (the middle term is the geometric mean of the outer two).

Here $a=-\dfrac27$, $b=x$, $c=-\dfrac72$, so:

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