Sequences & Series
Sequences and Series
nta_pyq_2025_jan
Grade 11
Question:
In an arithmetic progression, if $S_{40}=1030$ and $S_{12}=57$, then $S_{30}-S_{10}$ is equal to:
Step-by-Step Solution
Key Concept: Two equations $S_{n}=\tfrac{n}{2}(2a+(n-1)d)$ give $a$ and $d$. Then $S_{30}-S_{10}=5(4a+78d)$ from direct subtraction.
From $S_{40}=1030$: $2a+39d=\dfrac{103}{2}.$ From $S_{12}=57$: $2a+11d=\dfrac{19}{2}.$
Subtract: $28d=42\Rightarrow d=\dfrac{3}{2}.$ Then $2a=\dfrac{19}{2}-\dfrac{33}{2}=-7\Rightarrow a=-\dfrac{7}{2}.$
$$S_{30}-S_{10}=15(2a+29d)-5(2a+9d)=5(4a+78d)=5\bigl[-14+117\bigr]=515.$$
Correct Answer: 3