Question

1) if the sum of the cubes of first n natural numbers is 25502500, then find the value of n.2) how many terms are there in the sequence 3,21,147,........,50421?

Answer 1

Answer:

1) n = 100

2) There are 6 terms

Step-by-step explanation:

1) Let S be the sum of the first n cubes .

The formula for getting S is :

S = {n(n + 1)/2}²

We substitute the sum in this formula as follows :

25502500 = {n(n + 1)/2}²

We get the square root of both sides as follows:

√25502500 = √{n(n + 1)/2}²

5050 = n(n + 1)/2

Multiplying both sides by 2 we have:

10100 = n(n + 1)

10100 = n² + n

This forms a quadratic equation.

We solve using the quadratic formula:

n = {-1 +/-√1² + 4 × 1 × 10100}/2

n =( -1 +/-201)/2

n = 100 or  -101

We take the positive value of n and hence :

n = 100

2) This is a geometric progression.

Let the last term be n then.

aₙ = a₁rⁿ⁻¹

From the question :

aₙ = 50421

a₁ = 3

r = 21/3 = 7

Doing the substitution we have:

50421 = 3 × 7ⁿ⁻¹

50421/3 = 7ⁿ⁻¹

16807 = 7ⁿ⁻¹

Taking logs on both sides we have :

log 16807 = n-1 log7

n - 1 = log 16807/log 7

n - 1 = 5

n = 5 + 1

n = 6