Question

James wrote all 3 digit numbers which leaves a remainder 2 when divided by 3 Which is the last term

Answer 1

Answer:

The largest 3 digit number that will leave a remainder of 2 when divided by 3 is 998. So, it is an AP with the first term being 101 and the last term being 998 and common difference being 3.

Answer 2

• First 3 digit number which leaves 2 as remainder, when divided by 3, is 101
• Last 3 digit number which leaves 2 as remainder, when divided by 3, is 998

Here,

$\bf{}term_{ \tiny{ {1}^{st} }} = 101 \\ \boxed{ \bf{}term_{ \tiny{last }} = \red{998}} \\ \bf common \: \: difference = 3$

Now,

$\text{Number of terms(n)} = \rm{ \frac{a_{ \tiny{n}} - a_{ \tiny{1}}}{d} + 1} \\ \\ \implies \frac{998 - 101}{3} + 1 \\ \\ \implies \frac{897}{3} + 1 \\ \\ \implies299 + 1 \: \: \: = \sf \green{300}$