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Answer :

First term = a = 3,

Common difference = d = 8 – 3 = 13 – 8 = 5

\(a_n\) = 78

Using formula \(a_n = a + (n - 1)d\), to find \(n^{th}\) term of arithmetic progression,

\(\Rightarrow \) 78 = 3 + (n - 1) 5

\(\Rightarrow \) 75 = 5n - 5

\(\Rightarrow \) 80 = 5n

\(\Rightarrow \) n = 16

It means \(16^{th}\) term of the given AP is equal to 78.

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- Check whether –150 is a term of the AP: 11, 8, 5, 2…
- Find the 31st term of an AP whose \(11^{th}\) term is 38 and \(16^{th}\) term is 73.
- An AP consists of 50 terms of which \(3^{rd}\) term is 12 and the last term is 106. Find the \(29^{th}\) term.
- If the \(3^{rd}\) and the \(9^{th}\) terms of an AP are 4 and –8 respectively, which term of this AP is zero?
- The \(17^{th}\) term of an AP exceeds its \(10^{th}\) term by 7. Find the common difference.
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- Two AP’s have the same common difference. The difference between their \(100^{th}\) terms is 100, what is the difference between their \(1000^{th}\) terms.
- How many three digit numbers are divisible by 7?
- How many multiples of 4 lie between 10 and 250?
- For what value of n, are the nth terms of two AP’s: 63, 65, 67… and 3, 10, 17… equal?
- Determine the AP whose third term is 16 and the \(7^{th}\) term exceeds the \(5^{th}\) term by 12.
- Find the \(20^{th}\) term from the last term of the AP: 3, 8, 13… , 253.
- The sum of the \(4^{th}\) and \(8^{th}\) terms of an AP is 24 and the sum of \(6^{th}\) and \(10^{th}\) terms is 44. Find the three terms of the AP.
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- Ramkali saved Rs. 5 in the first week of a year and then increased her weekly savings by Rs. 1.75. If in the nth week, her weekly savings become Rs 20.75, find n.

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