Question

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

Answer 1

Answer:

304

Step-by-step explanation:

Given:

• Sum of first four terms of an A.P = 28
• Sum of first eight terms of the A.P = 88

To find:

• Sum of first 16 terms of the A.P

Formula:

Sum of n terms of an A.P = $\frac{n}{2}(2a+(n-1)d)$

Substituting for sum of first four terms:

$\frac{4}{2}(2a+(n-1)d)=28$

$2(2a+(n-1)d)=28$

$(2a+(4-1)d)=\frac{28}{2}$

$(2a+3d)=14$ ------(1)

Now substituting for the sum of first eight terms:

$\frac{8}{2} (2a+(8-1)d) = 88$

$4(2a+7d)=88$

$(2a+7d)=\frac{88}{4}$

$(2a+7d)=22$-------(2)

(2)-(1):

4d = 22-14

4d=8

$d=\frac{8}{4}$

d=2

The common difference is equal to 2

Now for finding a, let us substitute d in (2):

(2a+7×2)=22

2a=22-14

2a=8

a=4

The value of first term is 4

Sum of first 16 terms of A.P :

$\frac{16}{2}(2 \times 4+(16-1) 2 )$

$8(8+30)$

8(38)

=304

The sum of first 16 terms of the A.P is equal to 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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