Q.Match the entries in Column I with Column II. Column I:

We start with two unknown positive quantities; call them $a$ and $b$.

The arithmetic mean is given as $k = \frac{a+b}{2}$.

The geometric means between $a$ and $b$ — here “geometric means” means the two numbers that, together with $a$ and $b$, form a geometric progression. In a geometric progression of four terms $a, p, q, b$, the ratios are equal: $\frac{p}{a} = \frac{q}{p} = \frac{b}{q}$.

Thus $p$ and $q$ are the two middle terms of a four-term geometric sequence.

From the equal ratios:

Also, the common ratio $r$ satisfies $p = ar$ and $q = ar^2$ and $b = ar^3$.

So $p = ar$, $q = ar^2$, and $b = ar^3$.

Now we express everything in terms of $a$ and $r$:

1. Arithmetic mean:

2. Geometric means:

3. Compute $p^3 + q^3$:

4. Relate to $k$ and $pq$: First, $pq = (ar)(ar^2) = a^2 r^3$. Also, from step 1, $a(1+r^3) = 2k$. …

We have a three-digit number. Let its digits be $a, b, c$ (hundreds, tens, units), so the number is $100a + 10b + c$. The digits taken in order are in geometric progression (GP), and if we add 1 to the middle digit they become an arithmetic progression (AP). Also, subtracting 594 gives the number with digits reversed: $100c + 10b + a$.

1. Set up the GP condition.

Since $a, b, c$ are in GP, we have

(All digits are integers from 0 to 9, and $a \neq 0$ because it's a three-digit number.)

2. Set up the AP condition after adding 1 to the middle digit.

The new triple is $a, b+1, c$. For an AP, the middle term is the average of the outer two:

Multiply by 2:

So

3. Use the reversal condition.

Subtracting 594 gives the reversed number:

Simplify:

Cancel $10b$ from both sides:

Bring terms together:

Divide by 99:

So the hundreds digit is 6 more than the units digit.

4. Solve the system of equations.

From (1): $a + c = 2b + 2$.

From (2): $a = c + 6$.

Substitute into (1):

Divide by 2:

Now use the GP condition $b^2 = a c$:

Substitute $a = c+6$ and $b = c+2$:

$(c+2)^2 = (c+6)c.$ …

We are told that $b$ is the arithmetic mean of $a$ and $c$. That gives a direct linear relation. Also, $(b-a), (c-b), a$ are in geometric progression, which gives a quadratic relation. Combining these lets us eliminate variables and find the ratio.

1. Use the arithmetic mean condition Since $b$ is the arithmetic mean of $a$ and $c$:

2. Use the geometric progression condition The three numbers $(b-a), (c-b), a$ are in GP. For three terms in GP, the square of the middle term equals the product of the first and third:

3. Substitute $c$ from step 1 into the GP equation First compute $c - b$:

So the GP condition becomes:

4. Solve the resulting equation Since $a, b, c$ are unequal, $b - a \neq 0$. Divide both sides by $(b - a)$:

5. Find $c$ in terms of $a$ From $c = 2b - a$ and $b = 2a$: $$$$

Step 1 — Line up the terms with their index.

Step 2 — Numerators.

$1,3,5,7,\ldots$ is an AP with first term 1 and common difference 2:

Step 3 — Denominators.

$7^0,7^1,7^2,7^3,\ldots$ — the exponent is one less than the index:

Step 4 — Combine.

$T_n=\frac{2n-1}{7^{\,n-1}}$ …

A.P. condition: $2b=a+c$, so $b=\dfrac{a+c}{2}$.

G.P. condition: $b^2=ac$.

Substitute the first into the second: …

Concept: single AM and two GMs inserted between two positive numbers a and b.

x = AM = (a + b)/2.

For two GMs y, z: a, y, z, b are in GP with common ratio r, where b = a r^3 => r^3 = b/a.

y = a r, z = a r^2.

Numerator:

y^3 + z^3 = a^3 r^3 + a^3 r^6 = a^3 (b/a) + a^3 (b/a)^2 = a^2 b + a b^2 = ab(a + b).

Denominator: …

Step 1 — The general term of a GP.

For a geometric progression with first term $a_1 = a$ and common ratio $r$,

Step 2 — Use the given ratio.

The first term $a$ cancels — which is why we never need its value. We are told this equals 25, so

Step 3 — Express the required ratio the same way.

Step 4 — Substitute.

Write $r^4$ in terms of the quantity we actually know:

And since $25 = 5^2$,

Step 5 — Note why we never solve for $r$. …

Concept: use the AP condition to write a common difference, then apply the GP condition.

a, b, c in AP => b - a = c - b = d (common difference), so b = a + d and c = a + 2d.

The numbers (b - a), (c - b), a are in GP, i.e. d, d, a are in GP.

GP condition: (middle)^2 = product of the extremes

=> d^2 = d * a

=> d = a (d cannot be 0, else the GP is degenerate). …

The key idea is that the sum of $n$ terms of an A.P. is a quadratic in $n$ with no constant term: $S_n = \frac{n}{2}[2a + (n-1)d]$. When expanded, this becomes $\frac{d}{2}n^2 + (a - \frac{d}{2})n$. So if you are given $S_n = n^2 + n$, you can directly compare coefficients to find $d$ — but let's do it step by step to build understanding.

1. Find the $n$th term from the sum. For any sequence, the $n$th term $T_n$ is the difference between the sum of $n$ terms and the sum of $n-1$ terms:

Here $S_n = n^2 + n$, so $S_{n-1} = (n-1)^2 + (n-1)$.

Expand $S_{n-1}$:

Therefore:

2. Interpret $T_n$.

   The $n$th term of the A.P. is $T_n = 2n$. This is a linear expression in $n$, which is exactly what we expect for an A.P. — the general term is $a + (n-1)d$.

   For $n=1$, $T_1 = 2(1) = 2$, so the first term $a = 2$.

   For $n=2$, $T_2 = 2(2) = 4$, so the second term is $4$.

3. Find the common difference.

   The common difference $d$ is the difference between any two consecutive terms:

   $d = T_2 - T_1 = 4 - 2 = 2$ …