Arithmetic Progressions: CBSE Class 10 Mathematics practice questions

C. 32 cm

Here $a = 60$, $d = -2$ and $n = 15$, so $a_n = a + (n-1)d = 60 + 14\times(-2) = 32$ cm. Using $nd$ instead of $(n-1)d$ gives $30$ cm, the 16th rung, because only $14$ steps are needed to reach the 15th rung. Using $d = +2$ gives $88$ cm, and $28$ is just $14d$ without adding $a$.

A. 27th

$a = 5$, $d = 7$, so $187 = 5 + (n-1)\times 7$, giving $n - 1 = \dfrac{182}{7} = 26$ and $n = 27$. Forgetting the $+1$ at the end gives $26$, while $28$ and $25$ come from slips in adding or subtracting $1$.

D. 158

$253 = 3 + (n-1)\times 5$ gives $n = 51$. The 20th term from the last is term number $51 - 20 + 1 = 32$ (not $31$), so $a_{32} = 3 + 31\times 5 = 158$. Alternatively reverse the list: $a = 253$, $d = -5$, and $a_{20} = 253 - 19\times 5 = 158$. The value $153$ is the 21st from the last, $98$ is the 20th from the start, and $148$ is the 22nd from the last.