Ex 5.2 Q16 - Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.
Solution:- First term = a
Common difference = d
➙ Given that, the third term of the AP is 16,
∴ a₃ = 16
*Using formula, aₙ = a + (n - 1)d
∴ a + (3 - 1)d = 16
∴ a + 2d = 16
∴ a = 16 - 2d ------- (i)
➙ Also, the 7th term of the AP exceeds the 5th term by 12,
∴ a₇ = a₅ + 12
*Using formula, aₙ = a + (n - 1)d
∴ a + (7 - 1)d = a + (5 - 1)d + 12
∴ a + 6d = a + 4d + 12
∴ 6d - 4d = 12
∴ 2d = 12
∴ d =
12
/
2
∴ d = 6
*Putting d = 6 in Eqⁿ(i)
∴ a = 16 - 2(6)
∴ a = 16 - 12
∴ a = 4
Therefore,
*First term
= a
= 4
*Second term, a₂
= a + d
= 4 + 6
= 10
*Third term, a₃
= a + 2d
= 4 + 2(6)
= 16
*So, the required AP is..
4, 10, 16...
Hence,
The AP is 4, 10, 16...
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Try This..
✒ Determine the AP if its 5th term is 19 and the 10th term is 39.
✒ If the 2nd term of an AP is 7 and the 8th term is 25, find the AP.
✒ Find the AP whose 4th term is 20 and the 6th term is 30.
❌ Common Mistakes
Misplacing terms:▸ Remember, the nth term is a + (n – 1)d, not just a + nd.Wrong subtraction:▸ When comparing two terms like the 7th and 5th, make sure you subtract correctly.Skipping simplification:▸ Always simplify your equations step-by-step to avoid confusion.Not verifying:▸ After finding a and d, plug them back into the term formula to double-check.
📝 Related Questions:-
- Ex 5.2 Q19 - Subba Rao started work in 1995 at an annual salary of ₹5000 and received an increment of ₹200 each year. In which year did his income reach ₹7000?
- Ex 5.2 Q20 - Ramkali saved ₹5 in the first week of a year and then increased her weekly savings by ₹1.75. If in the nth week, her weekly savings become ₹20.75, find n.
Queries Solved:-
Class 10 Ex 5.2
Ex 5.2 Q16 Class 10
Class 10 Ex 5.2 Q16
Class 10 Chap 5 Ex 5.2 Q16
Class 10 Arithmetic Progression
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