Question

Hence,
35.10
Type II ON FIDING THE NUMBER OF TERMS IN AN A.P. WHEN THEIR SUM IS GIVEN
EXAMPLE 11 How many terms of the series 54,51, 48,... be taken so that their sum is 513?Em
ICBSE
le double answer​

Answer 1

Answer:

18 terms will be the answer

Answer 2

$n =18 \:\:or \:\: 19$

EXPLAINATION

GIVEN

⇒ < a > = 54, 51, 48, .........

⇒ Sₙ = 513

FORMULA

If 'a' and 'd' be the first term and common difference of an AP, then sum upto n term of the AP is given by

$Sₙ = \frac{n}{2}[2a + (n-1)d]----(1)$

CALCULATION

Here, first term and common difference of the AP are

a = 54, d = -3

Substituting these values in (1), we get

$513 = \frac{n}{2}[2\times54 + (n-1)(-3)]$

$513 = \frac{n}{2}[108 -3n + 3]$

$513 = \frac{n}{2}[111 -3n]$

$513\times2 = n[111 -3n]$

$171\times2 = n[37 -n]$

$342 = 37n -n²$

$n² - 37n + 342 = 0$

$n =\frac{37±\sqrt{37² -4\times1\times342}}{2}$

$n =\frac{37±1}{2}$

$n =\frac{37+1}{2} \:\:or \:\: \frac{37-1}{2}$

$n =\frac{38}{2} \:\:or \:\: \frac{36}{2}$

$n =19 \:\:or \:\: 18$