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Arithmetic Progressions
EXERCISE 5.2
CBSE_NCERT_TEXTBOOK
Grade 10
Question:
Which term of the AP : 3, 15, 27, 39, . . . will be 132 more than its 54th term?
Step-by-Step Solution
Key Concept: Use the nth term formula of an arithmetic progression, $T_n = a + (n-1)d$, where $a$ is the first term and $d$ is the common difference. Equate the required term to the 54th term plus 132 and solve for $n$.
1. Identify the first term and common difference: \[ a = 3, \quad d = 15-3 = 12. \] 2. Write the general term of the AP: \[ T_n = a + (n-1)d = 3 + (n-1)\times12 = 12n - 9. \] 3. Find the 54th term using the formula: \[ T_{54} = 12\times54 - 9 = 648 - 9 = 639. \] 4. According to the problem, the required term $T_n$ satisfies: \[ T_n = T_{54} + 132. \] Substitute $T_{54}=639$: \[ T_n = 639 + 132 = 771. \] 5. Set the general expression equal to 771 and solve for $n$: \[ 12n - 9 = 771 \] \[ 12n = 771 + 9 = 780 \] \[ n = \frac{780}{12} = 65. \] 6. Hence the 65th term of the given AP is 132 more than its 54th term.
Answer: The 65th term.
Correct Answer:65
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