Question

10th term of an.Ap is 30 and 7th term is 18 find the 17th term?
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Answer 1

$\huge\underline\mathbb{\red Q\pink{U}\purple{ES} \blue{T} \orange{IO}\green{N :}}$

10th term of an.Ap is 30 and 7th term is 18 find the 17th term?

$\huge\underline\mathbb{\red S\pink{O}\purple{LU} \blue{T} \orange{IO}\green{N :}}$

★ Given that,

• a10 = 30
• a7 = 18

★ To find,

• a17 = ?

★ Let,

$\sf\: ◼ \: a_{10} = a + 9d = 30 ..... (i)$

$\sf\: ◼ \: a_{7} = a + 6d = 18 ..... (ii)$

From the equations (i) & (ii), we get

$\sf\: ⟹ 3d = 12$

$\sf\:⟹ d = \frac{12}{3}$

$\sf\:⟹ d = 4$

$\boxed{\bf{∴ d = 4}}$

• Substitute the value of d in (i) .

$\sf\:⟹ a + 9(4) = 30$

$\sf\:⟹ a + 36 = 30$

$\sf\:⟹ a = 30 - 36$

$\sf\:⟹ a = - 6$

★ Now,

$\sf\:↪ a_{17} = a + 16d$

• Substitute the values.

$\sf\:↪ a_{17}= -6 + 16(4)$

$\sf\:↪ a_{17}= -6 + 64$

$\sf\:↪ a_{17}= 58$

$\underline{\boxed{\bf{\purple{∴ Hence, 17^{th} \: term \: of \: AP \: is \: 58.}}}}</p><p>$

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★ Important formulae in AP :

$\sf\: ◼ a_{n} = a + (n - 1)d$

$\sf\:◼ \: S_{n} = \frac{n}{2} [ 2a + (n - 1)d ]$

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