.

Given series is 3 ×1² + 5 × 2² + 7 × 3² + …

It’s seen that,

n^th term, a_n = ( 2n + 1) n² = 2n³ + n²

Then, the sum of n terms of the series can be expressed as

Given series is 1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 + …

It’s seen that,

n^th term, a_n = n ( n + 1) ( n + 2)

= (n² + n) (n + 2)

= n³ + 3n² + 2n

Then, the sum of n terms of the series can be expressed as

Given series is 1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + …

It’s seen that,

n^th term, a_n = n ( n + 1)

Then, the sum of n terms of the series can be expressed as

Let’s consider the roots of the quadratic equation to be a and b.

Then, we have

We know that,

A quadratic equation can be formed as,

x² – x (Sum of roots) + (Product of roots) = 0

x² – x (a + b) + (ab) = 0

x² – 16x + 25 = 0 [Using (1) and (2)]

Therefore, the required quadratic equation is x² – 16x + 25 = 0

Given,

The amount deposited in the bank is Rs 500.

At the end of first year, amount = Rs 500(1 + 1/10) = Rs 500 (1.1)

At the end of 2^nd year, amount = Rs 500 (1.1) (1.1)

At the end of 3^rd year, amount = Rs 500 (1.1) (1.1) (1.1) and so on….

Therefore,

The amount at the end of 10 years = Rs 500 (1.1) (1.1) … (10 times)

= Rs 500(1.1)¹⁰

Given, the number of bacteria doubles every hour. Hence, the number of bacteria after every hour will form a G.P.

Here we have, a = 30 and r = 2

So, a₃ = ar² = (30) (2)² = 120

Thus, the number of bacteria at the end of 2^nd hour will be 120.

And, a₅ = ar⁴ = (30) (2)⁴ = 480

The number of bacteria at the end of 4^th hour will be 480.

a_n +1 = arⁿ = (30) 2ⁿ

Therefore, the number of bacteria at the end of n^th hour will be 30(2)ⁿ.

Given that A and G are A.M. and G.M. between two positive numbers.

And, let these two positive numbers be a and b.

Consider the two numbers be a and b.

Then, G.M. = vab.

From the question, we have

We know that,

The G. M. of a and b is given by vab.

Then from the question, we have

By squaring both sides, we get