Question

The 6th term of an ap is 5 times the first term and the 11th term exceeds twice the 5th term by 3 . Find the 8th term.

Answer 1

Answer:

$\huge{\red{\boxed{\boxed{\green{\sf{a8=33}}}}}}$

Step-by-step explanation:

$\huge{\red{\boxed{\boxed{\red{\underline{\green{\sf{SOLUTION-}}}}}}}}$

□a6=5(a1)

□a11=2(a5)+3

To find:

○a8=?

$a6 = 5(a) \\ = )a + 5d = 5a \\ = )4a - 5d = 0 - - - - - (1) \\ again \\ a11 = 2(a5) + 3 \\ = )a + 10d = 2(a + 4d) + 3 \\ = )a + 10d = 2a + 8d + 3 \\ = )2a - a + 8d - 10d = - 3 \\ = )a - 2d = - 3 - - - - - (2) \\ \\ multiplying \: 1 \: in \: (1) \: and \: 4 \: in \: (2) \\ then \: subtract \: (2) \: from (1) \\ = ) 4a - 5d - (4a - 8d) = 0 - ( - 12)\\ = ) - 5d + 8d = 12\\ = )3d = 12 \\ = )d = \frac{12}{3} \\ = )d = 4 \\ \\ putting \: value \: of \: d \: in \: (1) \\ = )4a - 5d = 0 \\ 4a - 5 \times 4 = 0 \\ = )4a - 20 \\ = )4a = 20 \\ = )a = 5 \\ \\ a8 = a + 7d \\ = )5 + 7 \times 4 \\ = )5 + 28 \\ = )33$

$\huge{\red{\boxed{\boxed{\green{\sf{a8=33}}}}}}$

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Answer 2

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