Question

How many terms of the series 2+4+6+...... amount to 42?​

Answer 1

Answer:

a = 2

d = 2

total sum = 42

n = ?

Total sum = n/2 ( 2a + (n - 1)d

42 = n/2 ( 2×2 + (n - 1)2

42 = n/2 (4 + 2n - 2 )

42 = n/2 (2n + 2)

42 = 2n^2/2 + 2n/2

42 = n^2 + n

n^2 + n - 42 = 0

n^2 + 7n - 6n - 42 = 0

n(n + 7) - 6(n + 7) = 0

(n + 7)(n - 6) = 0

here,

n = -7 or 6

Since the no. of terms can never be negatueve.

Therefore, n = 6 Ans...

Answer 2

$\large \pink{ \underline{ \underline{ \sf \: Answer : \: \: \: }}}$

$\star$Number of terms = 6

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

Given ,

AP : 2 + 4 + 6 + .......

Here ,

First term (a) = 2

Common difference (d) = 4 - 2 = 2

Sum (Sn) = 42

We know that , the sum of first n terms of an AP is given by :

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

$\sf \implies 42 = \frac{n}{2} \bigg (2 \times 2 + (n - 1) 2 \bigg ) \\ \\ \sf \implies 42= \frac{n}{2} (4 + 2n - 2) \\ \\ \sf \implies 42= \frac{n}{2} (2 + 2n) \\ \\ \sf \implies 42 = \frac{2n}{2} + \frac{2 {n}^{2} }{2} \\ \\ \sf \implies {n}^{2} + n -42 = 0 \\ \\ \sf \implies {n}^{2} + 7n - 6n - 42 = 0 \\ \\ \sf By \: middle \: term \: splitting \: method \\ \\ \sf \implies n(n + 7) - 6(n + 7) = 0 \\ \\ \sf \implies (n - 6)(n + 7) = 0 \\ \\ \sf \implies n = 6 \: or \: n = - 7 \: \: \{ ignore \: negative \: value\}$

Hence , the required value is 6

_______________ Boht Hard ❤