Question

find the last value of n such that 2+5+8+11.....to n trem less then 200​

Answer 1

Answer:

Let write nth term of series

$=a + (n - 1)d\\ \\ = 2 + (n - 1)3 \\ \\ = 3n - 1$

But given that

$3n - 1 < 200 \\ 3n < 201 \\ n < 67$

So

n = 66

Answer 2

Answer:

This is a question based on Arithmetic Progression.

Formulas to be applied:

$\bullet \:\:a_n = a + ( n -1 ) d$

According to the question, we are given with a series starting from 2 and going on till 'n' terms are obtained. Also a condition is given that, the value of 'n' is less than 200. This implies the term after n is greater than 200. Hence we are asked to find the value of n from this condition.

Series: 2, 5, 8, 11, ..., n

First Term ( a ) = 2

Common Difference = 3

Number of terms = n

Last term = ?

Applying in the formula we get,

→ aₙ = a + ( n - 1 ) d

→ aₙ = 2 + ( n - 1 ) 3

→ aₙ = 2 + 3n - 3

→ aₙ = 3n - 1

Applying the given condition, we get:

→ aₙ < 200

→ 3n - 1 < 200

→ 3n < 200 + 1

→ 3n < 201

→ n < 201 ÷ 3

→ n < 67

According to the condition n is not equal to 67. Therefore the maximum value of n is 66.