Question

If 10 terms is 137 and 9 terms is 126 then find 11th term.​

Answer 1

Explanation:

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$\tt \bull \: a_{10} = 137 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (given) \\ \\ \leadsto \tt \: a + (10 - 1)d = 137 \\ \\ \leadsto \tt \: \red{ a + 9d = 137} \: \: \green{ - - - (1)}$

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$\tt \bull\: a_{9} = 126\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (given) \\ \\ \leadsto \tt \: a + (9- 1)d = 126\\ \\ \leadsto \tt \: \red{ a + 8d = 126} \: \: \green{ - - - (2)}$

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Substrate eqn.(1) and eqn.(2),

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$\tt \cancel{a} + 9d = 137 \\ \tt \cancel{a }+ 8d = 126 \\ - - - - - - - \\ \tt \: \: \: \: \: \: \purple{d \: = \: 11}$

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Put d = 11 in eqn. (1),

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$\leadsto \tt \: a + 9 \times 11 = 137 \\ \\ \leadsto \tt \: a + 99 = 137 \\ \\ \leadsto \tt \: a = 137 - 99 \\ \\ \leadsto \tt \: \pink{a = 38}$

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Now, we ask to find 11th term of an AP,

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$\leadsto \tt \: a_{11} = 38 + (11 - 1) \times 11 \\ \\ \leadsto \tt \: a_{11} =38 + 10\times 11 \\ \\ \leadsto \tt a_{11} =38 + 110\\ \\ \leadsto \tt \orange{a_{11} =148}$

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Hence, 11th term of an AP is 148.

Answer 2

Step-by-step explanation:

please mark as brainlist answer........