Math, asked by samara1356, 1 year ago


4. The sum of how many terms of the A.P. 10, 12, 14, ... will be 190​

Answers

Answered by Anonymous
34

  \large \underline{ \underline{ \sf \: Solution : \:  \:  \: }}

Given ,

First term = 10

Common difference = 12 - 10 = 2

Sn = 190

We know that ,

  \large    \sf \pink{\fbox{\fbox{S_{n} =  \frac{n}{2} (2a + (n - 1)d)}}}

 \sf \implies 190 =  \frac{n}{2}  \bigg(2(10) + (n - 1)2 \bigg) \\  \\\sf \implies  380 = n(20 + 2n - 2) \\  \\\sf \implies  380= 18n + 2 {n}^{2}  \\  \\\sf \implies  2 {n}^{2}  + 18n - 380 = 0 \\  \\\sf \implies  2 {n}^{2}  + 38n - 20n  - 380 = 0 \\  \\\sf \implies  2n(n + 19) - 20(n + 19) = 0 \\  \\\sf \implies  (2n - 20)(n + 19) = 0 \\  \\\sf \implies   n = 10 \:  \: and \:  \: n =  - 19

  \sf \star \: \{ \: ignore \:the \: negative \: value \: of \: n  \: \}

Hence , the required value of n is 10

Answered by jayajinkya35
10

Given that :

Sum(sn)=190,

we know:

sn =  \frac{n}{2} (2a +  < n - 1 > d)

From above A.P. it's evident that:

d = 12 - 10

d = 2

a = 10

Now let's put this values into formula:

190 = \frac{n}{2}  (2 \times 10 +  < n - 1 >  \times 2)

 =  > 190 \times 2 = n(20 + 2n - 2)

 =  > 380  = n(18 + 2n)

 =  > 190 = 18n + 2n {}^{2}

Now, transpose all terms on one side:

 =  > 2n {}^{2}  + 18n - 380 = 0

Divide both the side by 2 we get;.

n {}^{2}  + 9n - 190 = 0

 =  > n {}^{2}  + 19n - 10n - 190 = 0

 =  > n(n + 19) - 10(n + 19)

 =  > n =  - 19 \: or \: n = 10

Since,for the number of terms negative are negotiable

therefore,

n = 10

So,

sum of first10 term of A.P.=190

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