Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12. 63
Step-by-Step Solution
Key Concept: Use the general term of an A.P., $T_n = a + (n-1)d$, to translate the given conditions into algebraic equations and solve for the first term $a$ and common difference $d$.
Let the first term of the A.P. be $a$ and the common difference be $d$.
1. Third term condition
\[ T_3 = a + 2d = 16 \]
This gives the first equation:
\[ a + 2d = 16 \] (Equation 1)
2. Difference between 7th and 5th terms
\[ T_7 - T_5 = (a + 6d) - (a + 4d) = 2d = 12 \]
Hence,
\[ 2d = 12 \]
\[ d = 6 \] (Equation 2)
3. Find the first term
Substitute $d = 6$ into Equation 1:
\[ a + 2(6) = 16 \]
\[ a + 12 = 16 \]
\[ a = 4 \]
4. Write the required A.P.
The A.P. is:
\[ \{a, a+d, a+2d, a+3d, \dots\} = \{4, 4+6, 4+2\times6, 4+3\times6, \dots\} = \{4, 10, 16, 22, 28, 34, 40, \dots\} \]
5. Verification
- Third term $= 4 + 2\times6 = 16$ (as given).
- 7th term $= 4 + 6\times6 = 40$, 5th term $= 4 + 4\times6 = 28$.
- Difference $= 40 - 28 = 12$, satisfying the second condition.
Thus the required A.P. has first term $a = 4$ and common difference $d = 6$.
Correct Answer: First term $a = 4$, common difference $d = 6$; the A.P. is $4, 10, 16, 22, 28, 34, 40, \dots$.