Question

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Answer 1

a) 25

c) n

d) 2020

Step-by-step explanation:

Basically All question is related to Arithmetic Sequence, It's mean that here are terms with common difference.

Solution :

a) and b)

Here, First Term(a) = 1

common difference(d) = 3 - 1 = 5 - 3 = 2

As we know,

$\sf \: \text{Sum to } \: \blue{\bold{n}} \: terms \: ({S}_n) = \dfrac{n}{2} \{2a + (n - 1)d\}$

So, Sum for 5 terms

$= \sf \: \dfrac{5}{2} \{2.1 + (5 - 1)2\} \\ \\ \frac{5}{2} (2 + 4.2) \\ \\ = \frac{5}{2} .10 \\ \\ = 25$

(a) Therefore, Our required sum to 5 terms is 25

(b) and n terms is

$\sf {S}_n = \dfrac{n}{2} \{2a + (n - 1)d\}$

c)

a = 1/n

d = 3/n-1/n = 5/n - 3/n = 2/n

Now,

$\sf \: {S}_y = \dfrac{y}{2} \{2a + (y- 1)d\} \\ \\ \sf= \frac{n}{2} \{2. \frac{1}{n} + (n - 1) \frac{2}{n} \} \\ \\ \sf = \frac{n}{2} ( \frac{2}{n} + n. \frac{2}{n} - \frac{2}{n}) \\ \\ \sf = \frac{n}{2} ( \cancel{\frac{2}{n}} + \cancel{n}. \frac{2}{\cancel{n}} - \cancel{\frac{2}{n}}) \\ \\ = \sf \: \frac{n}{2} \times 2 = n$

Therefore, Sum of first n terms is n

d)

a = 1/2020

d = 3/2020 - 1/2020 = 5/2020 - 3/2020

Now,

Sum of 2020 terms,

$\sf {S}_n = \dfrac{n}{2} \{2a + (n - 1)d\} \\ \\ \sf {S}_{2020}= \dfrac{2020}{2} \{2. \frac{1}{2020} + (2020- 1) \frac{2}{2020} \} \\ \\ \sf \: = \frac{2020}{2} ( \frac{2}{2020} + 2020. \frac{2}{2020} - \frac{2}{2020} ) \\ \\ = \frac{2020}{2} (2) \\ \\ = 2020$

Therefore, Sum of first 2020 terms are 2020