Question

sum of 10terms of 1,2/3,4/9 is
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Answer 1

Answer:

a=1

d=2/3-1=-1/3

$s10 = \frac{10}{2} (2a + (n - 1)d) \\ = \frac{10}{2} (2 \times 1 + (10 - 1) \frac{ - 1}{3} ) \\ = 5(2 + 9 \times \frac{ - 1}{3} ) \\ = 5(2 + ( - 3)) \\ = 5( - 1) \\ = - 5$

Answer 2

Answer:

174075/59049

Step-by-step explanation:

Given,

1 , 2/3 , 4/9

To Find :-

Sum of 10 terms of 1 , 2/3 , 4/9.

How To Do :-

As they not said that terms are in A.P , G.P , H.P . So first we need to find the that the terms are in which progression. After finding that progression we can come to know about the progression and we can use that formula of sum of 'n' terms of that progression.

Points to be remembered :-

1) If a , b , c are in A.P

If and only if their common difference should be same :-

b - a = c - b.

2) If a , b , c are in G.P

If and only if their ratio should be same :-

b/a = c/b.

3) If a , b , c are in H.P

If and only if the difference of reciprocal of the terms are same

1/b - 1/a = 1/c - 1/b.

Solution :-

1 , 2/3 , 4/9

Checking whether the terms are in A.P

2/3 - 1 = 4/9 - 2/3

(2 - 3)/3 = (4 - (2 × 3))/9

-1/3 = (4 - 6)/9

-1/3 ≠ -2/9

∴ Difference is not same.

→ 1 , 2/3 , 4/9 are not in A.P

Checking whether the terms are in G.P :-

$\dfrac{\dfrac{2}{3}}{1} =\dfrac{\dfrac{4}{9}}{\dfrac{2}{3}}$

2/3 × 1 = 4/9 × 3/2

2/3 = 2/3

∴ Ratio is same

→ 1 , 2/3 , 4/9 are in G.P.

Sum of first 'n' terms of a G.P :-

$S_n=\dfrac{a(r^n-1)}{r-1}\:If,r>1$

$S_n=\dfrac{a(1-r^n)}{1-r}\:if\:r<1$

As we already obtained the value of 'r' = 2/3

∴ r = 2/3 < 1

So we need to use the second formula :-

Here :-

a = 1

n = 10

r = 2/3

S_10 = 1(1 - (2/3)¹⁰)/1 - 2/3

$=\dfrac{1-\dfrac{2^{10}}{3^{10}}}{\dfrac{(1\times3)-2}{3}}$

$=\dfrac{1-\dfrac{1024}{59049}}{\dfrac{3-2}{3}}$

$=\dfrac{\dfrac{59049-1024}{59049}}{\dfrac{1}{3}}$

$=\dfrac{\dfrac{58025}{59049}}{\dfrac{1}{3}}$

= 58025/59049 × 3/1

= 174075/59049