Question

The sixteen th term of an arithmetic sequence is 84 morethan its fourth term and 24 th term is 168 find first term and common difference

Answer 1

EXPLANATION.

• GIVEN

16th term of an Ap is 84 more than it's fourth term.

24 th term is = 168

FIND FIRST TERM AND COMMON DIFFERENCE.

According to the question,

16th term = 84 + 4th term

a + 15d = 84 + a + 3d .... (1)

12d = 84

d = 7

24 th term = 168

a + 23d = 168 ..... (2)

put the value of d =7 in equation (2)

we get,

a + 23(7) = 168

a + 161 = 168

a = 7

Therefore,

FIRST TERM = 7

COMMON DIFFERENCE = 7

Answer 2

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

The sixteen th term of an arithmetic sequence is 84 morethan its fourth term and 24 th term is 168 find first term and common difference.

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

★ Given that,

* 16th term of an AP is 84 more than 4th term.

* 24th term of an AP is 168.

★ To find,

• First term (a).
• Common difference (d).

★ Let,

$\sf\:⟹ a_{16} = 84 + a_{4}$

$\sf\:⟹ a + 15d = 84 + a + 3d$

$\sf\:⟹ 15d - 3d = 84$

$\sf\:⟹ 12d = 84$

$\sf\:⟹ d = \frac {84}{12}$

$\sf\:⟹ d = 7$

★ Now,

$\sf\:⟹ a_{24} = 168$

$\sf\:⟹ a + 23d = 168$

• Substitute the value of d.

$\sf\:⟹ a + 23(7) = 168$

$\sf\:⟹ a + 161 = 168$

$\sf\:⟹ a = 168 - 161$

$\sf\:⟹ a = 7$

$\underline{\boxed{\bf{\red{ ∴ Hence,the \: value \: of \: first \: term \: is \: “ \: 7 \: ” , common \: difference (d) \: is \: “ \: 7 \: ”.}}}}\:\orange{\bigstar}$

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★ Important formulae in AP:

$\bf\:◼ \: a_{n} = a + (n - 1)d$

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

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