Question

30. Find the AP. for which 4th term is 15 and 8th term is one more than twice the 4th
term find fast​

Answer 1

Answer: in attachment

Step-by-step explanation:

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Answer 2

For an A.P.

According to first condition;

$a_{4} = 15 \\ </p><p>a + 3d = 15$ ... $(a_{n} = a + (n -1)d)$(i)

According to second condition;

$a_{8} = 1 + 2a_{4} \\ </p><p>a + 7d = 1 + 2(a + 3d)$ ... $a_{n} = a + (n-1)d)$

$a + 7d = 1 + 2a + 6d \\</p><p>2a - a + 1 = 7d - 6d \\</p><p>d = a + 1 \\ \\ substituting \: in \:(i), \:we \:get; \\ \\</p><p>a + 3a + 3 = 15 \\</p><p>4a = 12 \\</p><p>a = 3 \\</p><p>since d = a + 1 \\</p><p>d = 3 + 1 \\ </p><p>d = 4</p><p>$

Hence, the A.P. is 3, 7, 11, 15, 19, 23, ...