Question

How many terms of the A. P : 6,12,18, must be taken to give a sum 330​

Answer 1

Solution -

We have,

⠀⠀⠀➝ A.P. = 6, 12, 18, ...

⠀

We can see that the given series in an A.P. with

• First term (a) = 6
• Common difference (d) = 6

⠀

We have to find the number of terms that must be taken to give a sum 330. Therefore,

• $\sf{S_n = 330}$

⠀

Now, we use the formula given below :-

$\large{\bf{\longmapsto{\boxed{\pink{S_n = \dfrac{n}{2}[2a + (n -1)d]}}}}}$

⠀

Putting all the values in the formula

$\tt:\implies\: \: \: \: \: \: \: \: {330 = \dfrac{n}{2} [2(6) + (n -1)6]}$

⠀

$\tt:\implies\: \: \: \: \: \: \: \: {330 \times 2 = n[12 + 6n - 6]}$

⠀

$\tt:\implies\: \: \: \: \: \: \: \: {660 = n[6n + 6]}$

⠀

$\tt:\implies\: \: \: \: \: \: \: \: {660 = 6n^2 + 6n}$

⠀

$\tt:\implies\: \: \: \: \: \: \: \: {6n^2 + 6n - 660 = 0}$

⠀

Taking 6 as common

$\tt:\implies\: \: \: \: \: \: \: \: {6(n^2 + n - 110) = 0}$

⠀

$\tt:\implies\: \: \: \: \: \: \: \: {n^2 + n - 110 = 0}$

⠀

We can see that, it is in the form of quadratic equation.

Splitting the middle term

$\tt:\implies\: \: \: \: \: \: \: \: {n^2 + 11n - 10n - 110 = 0}$

⠀

$\tt:\implies\: \: \: \: \: \: \: \: {n(n + 11) - 10(n + 11) = 0}$

⠀

$\tt:\implies\: \: \: \: \: \: \: \: {(n + 11) (n - 10) = 0}$

⠀

We get,

➝ n = -11⠀⠀⠀❲✘❳

➝ n = 10⠀⠀⠀❲✔❳

[Since, number of terms cannot be negative]

⠀

★ $\small\underline{\sf{Hence,\: number\: of\: terms\: must\: be\: taken\: is\: 10.}}$

Answer 2

Answer:

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