Math, asked by challa2418, 1 year ago

Find the least value of n for which 1 + 7 + 7^2 + ... + 7^n > 1000.

Answers

Answered by marwabood1995
0

Answer:

Step-by-step explanation:

5n+7=7 (n+1)-2n

Simplifying

5n + 7 = 7(n + 1) + -2n

Reorder the terms:

7 + 5n = 7(n + 1) + -2n

Reorder the terms:

7 + 5n = 7(1 + n) + -2n

7 + 5n = (1 * 7 + n * 7) + -2n

7 + 5n = (7 + 7n) + -2n

Combine like terms: 7n + -2n = 5n

7 + 5n = 7 + 5n

Add '-7' to each side of the equation.

7 + -7 + 5n = 7 + -7 + 5n

Combine like terms: 7 + -7 = 0

0 + 5n = 7 + -7 + 5n

5n = 7 + -7 + 5n

Combine like terms: 7 + -7 = 0

5n = 0 + 5n

5n = 5n

Add '-5n' to each side of the equation.

5n + -5n = 5n + -5n

Combine like terms: 5n + -5n = 0

0 = 5n + -5n

Combine like terms: 5n + -5n = 0

0 = 0

Solving

0 = 0

Answered by ushmagaur
0

Answer:

The least value of n is 5.

Step-by-step explanation:

Concept:-

Sum of n terms of GP,

S_n=\frac{a(r^n-1)}{r-1}

Step 1 of 2

Consider the given inequality as follows:

1+7+7^2+...+7^n > 1000 . . . . . (1)

Rewrite the inequality (1) as follows:

7^0+7^1+7^2+...+7^n > 1000 . . . . . (2)

Notice that the left-hand side series is a geometric progression (GP) on n terms such that the first term, a = 7^0

and common ratio, r = 7.

Simplify the left-hand side as follows:

S_n=\frac{7^0(7^n-1)}{7-1}

    =\frac{1(7^n-1)}{6}

Thus, the inequality (2) becomes,

\frac{(7^n-1)}{6} > 1000

Multiply both the sides by the number 6 as follows:

6\times \left(\frac{7^n-1}{6}\right) > 6\times 1000

7^n-1 > 6000

Add the number 1 both the sides as follows:

7^n-1 +1 > 6000+1

7^n > 6001 . . . . (3)

Step 2 of 2

Substituting the values for n in (3) to get the least value for n.

For n=1,

7^1 > 6001

⇒ 7 > 6001, which is not possible.

For n=2,

7^2 > 6001

⇒ 49 > 6001, which is not possible.

For n=3,

7^3 > 6001

⇒ 343 > 6001, which is not possible.

For n=4,

7^4 > 6001

⇒ 2401 > 6001, which is not possible.

For n=5,

7^5 > 6001

⇒ 16807 > 6001, which is true.

Therefore, the least value of n for which the given inequality holds is 5.

#SPJ2

Similar questions