QUESTION 1
Easy
Find the \(31^{st}\) term of the AP whose \(11^{th}\) term is 38 and \(16^{th}\) term is 73.
SOLUTION
1
Write down what is given
\(T_{11} = 38\)
⇒ \(a + 10d = 38\) ---(I)
and \(T_{16} = 73\)
⇒ \(a + 15d = 73\) ---(II)
⇒ \(a + 10d = 38\) ---(I)
and \(T_{16} = 73\)
⇒ \(a + 15d = 73\) ---(II)
2
Find the common difference
Subtracting (I) from (II), we get
\(a + 10d − a − 15d = 73 − 38\)
⇒ \(5d = 35\)
⇒ \(d = 7\)
\(a + 10d − a − 15d = 73 − 38\)
⇒ \(5d = 35\)
⇒ \(d = 7\)
3
Find the first term
Substituting \(d = 7\) in (I), we get
\(a + 10(7) = 38\)
⇒ \(a + 70 = 38\)
⇒ \(a = −32\)
\(a + 10(7) = 38\)
⇒ \(a + 70 = 38\)
⇒ \(a = −32\)
4
Find the \(31^{st}\) term
\(\begin{aligned} T_{31} &= a + 30d \\ &= −32 + 30(7) \\ &= − 32 + 210 \\ &= 178 \end{aligned}\)
🏆
Final Answer : \(T_{31} = 178\)
Concept Note
When two terms of an AP are known,
1. Use \(T_n = a + (n − 1)d\) to form equations.
2. Solve for \(a\) and \(d\).
3. Substitute into the formula to find any required term.