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प्रश्न
Find the first terms and common difference of an A.P. whose t8 = 3 and t12 = 52.
उत्तर
For an A.P., let a be the first term and d be the common difference.
t8 = 3 and t12 = 52 ......[Given]
Since tn = a + (n – 1)d,
t8 = a + (8 – 1)d
∴ 3 = a + 7d
i.e., a + 7d = 3 ......(i)
Also, t12 = 52
∴ a + (12 – 1)d = 52
∴ a + 11d = 52 ......(ii)
Subtracting equation (i) from (ii), we get
a + 11d = 52
a + 7d = 3
– – –
4d = 49
∴ d = `49/4`
Substituting d = `49/4` in equation (i), we get
`"a" + 7(49/4)` = 3
∴ `"a" + 343/4` = 3
∴ a = `3 - 343/4`
= `(-331)/4`
∴ The first term and common difference of A.P. are `(-331)/4` and `49/4` respectively.
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| Treasure Hunt is an exciting and adventurous game where participants follow a series of clues/numbers/maps to discover hidden treasure. Players engage in a thrilling quest, solving puzzles and riddles to unveil the location of the coveted prize. While playing a treasure hunt game, some clues (numbers) are hidden in various spots collectively forming an A.P. If the number on the nth spot is 20 + 4n, then answer the following questions to help the players in spotting the clues: |
- Which number is on first spot? 1
- Which spot is numbered as 112? 2
OR - What is the sum of all the numbers on the first 10 spots? 2
- Which spot is numbered as 112? 2
- Which number is on the (n – 2)th spot? 1
