Question

If the sum of first four terms of an A.P. is 28 and sum of the first eight terms of the same A.P. is 88, then sum of first 16 terms of the A.P is

Answer 1

Answer:-

Given:

sum of first 4 terms of an AP = 28.

sum of first 8 terms = 88

We know that,

Sum of first n terms of an AP = n/2 * [ 2a + (n - 1)d ]

Hence,

→ S(4) = 28

→ 4/2 [ 2a + (4 - 1)d ] = 28

→ 2 (2a + 3d) = 28

→ 2a + 3d = 28/2

→ 2a + 3d = 14 -- equation (1)

Similarly,

→ S(8) = 88

→ 8/2 * [ 2a + (8 - 1)d ] = 88

→ 4 (2a + 7d) = 88

→ 2a + 7d = 88/4

→ 2a + 7d = 22 -- equation (2)

On subtracting equation (1) from (2) we get,

→ 2a + 7d - (2a + 3d) = 22 - 14

→ 2a + 7d - 2a - 3d = 8

→ 4d = 8

→ d = 8/4

→ d = 2

Putting the value of d in equation (1) we get,

→ 2a + 3d = 14

→ 2a + 3*2 = 14

→ 2a = 14 - 6

→ 2a = 8

→ a = 8/2

→ a = 4

Hence,

→ S(16) = 16/2 [ 2 * 4 + (16 - 1)(2) ]

→ S(16) = 8 (8 + 30)

→ S(16) = 8 (38)

→ S(16) = 304

Therefore, the sum of first 16 terms of the given AP is 304.

Answer 2

$\bf\huge\blue{\underline{\underline{ Question : }}}$

If the sum of first four terms of an A.P. is 28 and sum of the first eight terms of the same A.P. is 88, then sum of first 16 terms of the A.P is

$\bf\huge\blue{\underline{\underline{ Solution : }}}$

★ Given that,

$\sf\:\rightarrow Sum\:of\:first\:four\:terms\:of\:an\:AP = 28$

• $\tt\:\ S_{4} = 28$

$\sf\:\rightarrow Sum\:of\:first\:eight\:terms\:of\:an\:AP = 88$

• $\tt\: S_{8} = 88$

★ To find,

$\bf\:\rightarrow Sum\:of\:first\:sixteen\:terms\:of\:an\;AP.$

★ Formula :

$\boxed{\rm{\red{\bigstar \: S_{n} = \frac{n}{2} [ 2a + (n - 1)d ] }}}$

★ Let,

➡ ᴄᴀsᴇ - 1 :-

• S8 = 28
• n = 4

$\sf\:\implies 28 = \cfrac{4}{2} [ 2a + (4 - 1)d ]$

$\sf\:\implies 28 = 2 [ 2a + 3d ]$

$\sf\:\implies \cfrac{28}{2}= 2a + 3d$

$\sf\:\implies 2a + 7d = 14 ..... (1)$

➡ ᴄᴀsᴇ - 2 :-

• S8 = 88
• n = 8

$\sf\:\implies 88 = \cfrac{8}{2} [2a + (8-1)d]$

$\sf\:\implies 88 = 4[2a + 7d]$

$\sf\:\implies \cfrac{88}{4} =2a + 7d$

$\sf\:\implies 2a + 7d = 22.....(2)$

★ From,

Subtract equations (1) & (2). We get,

$\sf\:\implies - 4d = - 8$

$\sf\:\implies d = \cfrac{- 8}{-4}$

$\sf\:\implies d = 2$

• Substitute value of d in (1).

$\sf\:\implies 2a + 3(2) = 14$

$\sf\:\implies 2a + 6 = 14$

$\sf\:\implies 2a = 14 - 6$

$\sf\:\implies 2a = 8$

$\sf\:\implies a = \cfrac{8}{2}$

$\sf\:\implies a = 4$

★ Verification,

Verify whether these values are correct or not.

• Substitute values of a & d in (1), to get LHS = RHS.

LHS =

$\sf\:\implies 2(4)+3(2)$

$\sf\:\implies 8 + 6$

$\sf\:\implies 14$

◼ Since, LHS = RHS.

◼ Hence, it was verified.

★ Now,

• We can find out the value of sum of first 16 terms of an AP.

$\sf\:\implies S_{16} = \cfrac{16}{2} [2(4)+(16-1)2]$

$\sf\:\implies S_{16} = 8[8+(15)2]$

$\sf\:\implies S_{16} = 8[8+30]$

$\sf\:\implies S_{16} = 8[38]$

$\sf\:\implies S_{16} = 304$

$\underline{\boxed{\rm{\purple{\therefore Sum\:of\:first\:16\:terms\:of\:an\:AP\:is\:304.}}}}\:\orange{\bigstar}$

★ More information :

$\boxed{\begin{minipage}{5 cm} AP Formulae : \\ \\ : \implies a_{n} = a + (n - 1)d \\ \\ :\implies S_{n} = \frac{n}{2} [ 2a + (n - 1)d ] \end{minipage}}$

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