Exercise 5.2: Arithmetic Progressions — NCERT Solutions

Here $a = 2$, $d = 5 - 2 = 3$, and $n = 12$.

Using $a_n = a + (n-1)d$: $a_{12} = 2 + (12 - 1)\times 3 = 2 + 11\times 3 = 2 + 33 = 35$

The 12th term is $35$.

Here $a = 3$, $d = 5$, and we want $a_n = 78$.

$a + (n-1)d = 78$ $3 + (n-1)\times 5 = 78$ $(n-1)\times 5 = 75$ $n - 1 = 15 \implies n = 16$

So $78$ is the 16th term.

Using $a_n = a + (n-1)d$: $a_7 = a + 6d = 34 \quad (1)$ $a_{13} = a + 12d = 64 \quad (2)$

Subtract (1) from (2): $6d = 30 \implies d = 5$

Substitute in (1): $a + 6(5) = 34 \implies a = 34 - 30 = 4$

The AP is $4, 9, 14, 19, \dots$

Here $a = 11$, $d = 8 - 11 = -3$.

Suppose $a_n = 150$: $11 + (n-1)(-3) = 150$ $(n-1)(-3) = 139$ $n - 1 = -\frac{139}{3}$

This is not a positive integer, so $150$ is not a term of the AP. (Also note this AP decreases, so it never reaches 150.)

The two-digit multiples of 3 are $12, 15, 18, \dots, 99$.

This is an AP with $a = 12$, $d = 3$, last term $a_n = 99$.

$a + (n-1)d = 99$ $12 + (n-1)3 = 99$ $(n-1)3 = 87 \implies n - 1 = 29 \implies n = 30$

There are 30 two-digit numbers divisible by 3.