Question

11,22,33. In this arthematic sequence Find 30 th term. Find algebraic form. Find sum if first 25 terms.





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Answer 1

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a = 11

d = 22 - 11 = 11

n = 25

$sn \: = \frac{n}{2} \times (2a \: + ( \: n \: - \: 1 \: )d \: ) \:$

$sn \: = \: \frac{25 }{2} \times( 2 \: \times 11 + (25 - 1)11)$

$sn \: = \frac{25}{2} \times (22 + 24 \times 11)$

$sn \: = \frac{25}{2} (22 + 264)$

$sn = \frac{25}{2} \times (286)$

$sn \: = 25 \times 143$

sn = 3575

so the sum of first 25 terms of arithmetic sequence is 3575

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Answer 2

Answer:

• 30th term = 330

• nth term = 11n

• sum of first 25 terms = 3575

Step-by-step explanation:

Given :

Arithmetic sequence : 11 , 22 , 33

To find :

• 30th term
• algebraic form
• sum of first 25 terms

Solution :

In the given Arithmetic sequence,

first term, a = 11

common difference, d = 22 - 11 = 11

• nth term of Arithmetic Progression is given by,

 $\longmapsto \bf a_n=a+(n-1)d$

To find the 30th term, put n = 30

a₃₀ = a + (30 - 1)d

a₃₀ = 11 + 29(11)

a₃₀ = 11 + 319

a₃₀ = 330

∴ 30th term = 330

• Now, algebraic form of nth term

aₙ = a + (n - 1)d

aₙ = 11 + (n - 1)(11)

aₙ = 11 + 11n - 11

aₙ = 11n

∴ nth term of given A.P. = 11n

• Sum of first n terms is given by,

$\longmapsto \bf S_n=\dfrac{n}{2}[2a+(n-1)d]$

Put n = 25 to find the sum of first 25 terms,

$\sf S_{25}=\dfrac{25}{2}[2(11)+(25-1)(11)] \\\\ \sf S_{25}=\dfrac{25}{2}[22+24(11)] \\\\ \sf S_{25}=\dfrac{25}{2}[22+264] \\\\ \sf S_{25}=\dfrac{25}{2}[286] \\\\ \sf S_{25}=25 \times 143 \\\\ \sf S_{25}=3575$

∴ The required sum of first 25 terms is 3575