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Answer 1

QUESTION:

Let S = 3 + 55 + 333 + 5555 + 33333 + ........upto 22 terms. If 9S + 88 = A(10²² - 1), then A is

ANSWER:

• OPTION 2) 530/99

GIVEN:

• S = 3 + 55 + 333 + 5555 + 33333 + ........upto 22 terms

• 9S + 88 = A(10²² - 1)

TO FIND:

• The value of A.

EXPLANATION:

Split the two terms

S = (3 + 333 + 33333 .... upto 11 terms) + (55 + 5555 + 555555 + .......upto 11 terms)

Let 3 + 333 + 33333 .... upto 11 terms = P

Let 55 + 5555 + 555555 + .......upto 11 terms = Q

S = P + Q

P = 3 + 333 + 33333 .... upto 11 terms  

Take 3 as common

P = 3(1 + 111 + 11111 + ....11 terms)

Multiply and divide by 9

P = 3/9(9 + 999 + 99999 + ....11 terms)

P = 3/9(10 - 1 + 1000 - 1  + 100000 - 1  + ....11 terms)

Again split the terms  

P = 3/9[(10 + 1000 + 100000 + ....11 terms) - ( 1 + 1 + 1  + .... 11 terms)]

(10 + 1000 + 100000 + ....11 terms) is in G.P.

$\boxed{\bold{\large{S_n = \dfrac{a(r^n - 1)}{r-1}}}}$

r = t₂ / t₁ = 1000 / 10 = 100

a = 10

n = 11

S₁₁ =  10(100¹¹ -  1) / 100 - 1

100¹¹ =  10²² as 100 = 10²

S₁₁ =  10(10²² -  1) / 99

(1 + 1 + 1  + .... 11 terms) are also in G.P.

All the 11 terms are same  

$\boxed{\bold{\large{S_n = na}}}$

a = 1

n = 11

$S_n = 11$

Substitute the values which we found.

P = 3/9[(10(10²² -  1) / 99) - 11]

By taking L.C.M

P = 3/891[(10(10²² -  1) - 1089]

Q = 55 + 5555 + 555555 .... upto 11 terms  

Take 5 as common

Q = 5(11 + 1111 + 111111 + ....11 terms)

Multiply and divide by 9

Q = 5/9(99 + 9999 + 999999 + ....11 terms)

Q = 5/9(100 - 1 + 10000 - 1  + 1000000 - 1  + ....11 terms)

Again split the terms  

Q = 5/9[(100 + 10000 + 1000000 + ....11 terms) - ( 1 + 1 + 1  + .... 11 terms)]

(100 + 10000 + 1000000 + ....11 terms) is in G.P.

$\boxed{\bold{\large{S_n={a(r^n- 1)}{r - 1}}}}$

r = t₂ / t₁ = 10000 / 100 = 100

a = 100

n = 11

S₁₁ =  100(100¹¹ -  1) / 100 - 1

100¹¹ =  10²² as 100 = 10²

S₁₁ =  100(10²² -  1) / 99

(1 + 1 + 1  + .... 11 terms) are also in G.P.

All the 11 terms are same

$\boxed{\bold{\large{S_n = na}}}$

a = 1

n = 11

$S_n = 11$

Substitute the values which we found

Q = 5/9[(10(10²² -  1) / 99) - 11]

By taking L.C.M

Q = 5/891[(100(10²² -  1) - 1089]

S = P + Q

P + Q = 3/891[(10(10²² -  1) - 1089] + 5/891[(100(10²² -  1) - 1089]

S = 1/891(30(10²² -  1) - 3267 + 500(10²² -  1) - 5445)

9/891(30(10²² -  1) + 500(10²²  -  1) - 8712) + 88 = A(10²² - 1)

1/99(30(10²² -  1) + 500(10²² -  1) - 8712) + 88 = A(10²² - 1)

Take (10²² -  1) as common.

1/99((10²² -  1) (30 + 500) - 8712) + 88 = A(10²² - 1)

1/99(530(10²² -  1) - 8712) + 88 = A(10²² - 1)

530/99(10²² -  1) - 8712/99 + 88 = A(10²² - 1)

530/99(10²² -  1) - 88 + 88 = A(10²² - 1)

530/99(10²² -  1) = A(10²² - 1)

Cancel (10²² -  1) on both sides.

 530/99 = A

HENCE THE VALUE OF A = 530/99.