Question

how many terms of the AP 3,5,7,9...must be added to get the sum 120?​

Answer 1

Answer:

Sn = n/2[2a+(n-1)d]

given,Sn=120,a=3 , d=2

120=n/2[2×3+(n-1)2]

120×2=n[6+2n-2]

240=n[4+2n]

240=4n+2n²

240=2(2n+n²)

120=n²+2n

0=n²+2n-120

0=n²+12n-10n-120

0=n(n+12)-10(n+12)

0=(n+12)(n-10)

n=-12 , 10

hence no of terms can't be negative

so n=10

thus there are 10 terms of the given AP must be added to get sum 120.

Step-by-step explanation:

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Answer 2

Given :-

• First term (a) = 3

• Common difference (d) = 5 - 3 = 2

To find :-

• Required number of terms

Solution :-

let the required number of terms be n,

We know that,

$\blue{\bigstar}\:\:{\underline{\boxed{\bf\red{s_n=\dfrac{n}{2}\:[2a+(n-1)d]}}}}$

$\rm\longrightarrow\:s_n=\dfrac{n}{2}\:[2\times{3}+(n-1)2]$

$\rm\longrightarrow\:s_n=\dfrac{n}{2}\:[6+(n-1)2]$

$\rm\longrightarrow\:s_n=\dfrac{n}{2}\:[6+2n-2]$

$\rm\longrightarrow\:s_n=\dfrac{n}{2}\:(2n+4)$

Now,

$\rm\longrightarrow\:\dfrac{n}{2}\:(2n+4)=120$

$\rm\longrightarrow\:n(2n+4)=120\times{2}$

$\rm\longrightarrow\:n(2n+4)=240$

$\rm\longrightarrow\:{2n}^{2}+4n=240$

$\rm\longrightarrow\:2({n}^{2}+2n)=240$

$\rm\longrightarrow\:{n}^{2}+2n=\cancel\dfrac{240}{2}$

$\rm\longrightarrow\:{n}^{2}+2n=120$

$\rm\longrightarrow\:{n}^{2}+2n-120=0$

$\rm\longrightarrow\:{n}^{2}+12n-10n-120=0$

$\rm\longrightarrow\:n(n+12)-10(n+12)=0$

$\rm\longrightarrow\:(n+12)(n-10)=0$

Now,

$\rm\:n+12=0$

$\rm\longrightarrow\:n=0-12$

$\rm\longrightarrow\:n=-12$

Then,

$\rm\:n-10=0$

$\rm\longrightarrow\:n=0+10$

$\rm\longrightarrow\:n=10$

In the above Solution,

n ≠ -12

So,here n is only 10

Hence,the required number of terms will be 10.