Question

how many terms of arthematic sequence 99, 97, 95 must be added to get 900?​

Answer 1

Solution :-

→ First term of AP = 99 = a

→ common difference = 97 - 99 = (-2) = d

→ Let number of terms added = n

→ sum = Sn = 900

Putting all values in AP sum formula :-

• Sn = (n/2)[2a + (n - 1)d]

we get :-

→ 900 = (n/2)[2*99 + (n-1)(-2)]

→ 1800 = n(198 - 2n + 2)

→ 1800 = 200n - 2n²

→ 2n² - 200n + 1800 = 0

→ n² - 100n + 900 = 0

→ n² - 90n - 10n + 900 = 0

→ n(n - 90) - 10(n - 90) = 0

→ (n - 90)(n - 10) = 0

→ n = 90 or 10 . (Ans.)

Hence, 90 and 10 terms of AP gives sum 900.

Answer 2

★ Given that,

• how many terms of arithmetic sequence 99, 97, 95 must be added to get 900?

★ Let,

• a = 99
• a2 = 97

➡ Common difference = a2 - a = 97 - 99 = -2

• Sn = 900

★ Formula :

Sum of nth terms of an AP :

$\orange{\bigstar}\boxed{\rm{\red{ S_{n} = \cfrac{n}{2}[2a + (n-1) d]}}}\:\orange{\bigstar}$

• Substitute the values.

$\sf\:\implies 900 = \cfrac{n}{2} [ 2(99)+(n-1)(-2)]$

$\sf\:\implies 900 \times 2 = n[198 - 2n + 2]$

$\sf\:\implies 1800 = n[200 - 2n]$

$\sf\:\implies 1800 = 200n - 2n^{2}$

$\sf\:\implies 2n^{2} - 200n + 1800 = 0$

$\sf\:\implies 2(n^{2} - 100n + 900) = 0$

$\sf\:\implies n^{2} - 100n + 900= \cfrac{0}{2}$

$\sf\:\implies n^{2} - 100n + 900= 0$

• Let's factories it.

$\sf\:\implies n^{2} - 10n - 90n + 900 = 0$

$\sf\:\implies n(n- 10)- 90(n-10)= 0$

$\sf\:\implies (n- 10)(n-90)= 0$

$\sf\:\implies n = 10 (or) 90$

$\underline{\boxed{\rm{\purple{\therefore The\:terms\:are\:10\:and\:90.}}}}\:\orange{\bigstar}$