Question

The sum of the first 8 terms of an arithmetic sequence is 136 and the sum of the first 12 terms is 300. a) What is the sum of the first and the 8th terms? b) What is the sum of the first and the 12th terms ? c) What is the number got by adding three times the first term to the 19th term ?

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Answer 1

Answer:

34,  50,   84

Step-by-step explanation:

Let the first term be 'a' and common difference be 'd'.

           Given,      S₈ = 136

⇒ S₈ = (8/2) [2a + (8 - 1)d]

⇒ 136 = 4(2a + 7d)

⇒ 34 = 2a + 7d           ...(1)

            Given.      S₁₂ = 300

⇒ S₁₂ = (12/2) [2a + (12 - 1)d]

⇒ 300 = 6(2a + 11d)

⇒ 50 = 2a + 11d             ...(2)

a):  Sum of 1st and 8th term:

       ⇒ a + [ a + (8 - 1)d ]

       ⇒ 2a + 7d

       ⇒ 34                 [from (1)]

b):  Sum of 1st and 12th term:

       ⇒ a + [ a + (12 - 1)d]

       ⇒ 2a + 11d

       ⇒ 50                 [from (2)]

c): Number got by 3 times a and 19th term

       ⇒ 3*a + [a + (19 - 1)d ]

       ⇒ 4a + 18d

       ⇒ (2a + 11d) + (2a + 7d)

       ⇒ 50 + 34            [from (1) & (2)]

       ⇒ 84        

*Solved using

Sₙ = (n/2) [2a + (n - 1)d]      [sum]

aₙ = a + (n - 1)d       [nth term]

Answer 2

Question:

• The sum of the first 8 terms of an arithmetic sequence is 136 and the sum of the first 12 terms is 300. ❶ What is the sum of the first and the 8th terms? ❷ What is the sum of the first and the 12th terms? ❸ What is the number got by adding three times the first term to the 19th term?

Answer:

• ❶ Sum of first and 8th terms is 34.
• ❷ Sum of first and 12th term is 50.
• ❸ The number got by adding three times the first term to the 19th term is 84.

Explanation:

Given that:

• Sum of 1st 8 terms ($\sf S_{8}$) = 136
• Sum of 1st 12 terms ($\sf S_{8}$) = 300

To Find:

• ❶ Sum of first and 8th terms = ?
• ❷ Sum of first and 12th terms = ?
• ❸ The number got by adding three times the first term to the 19th term = ?

Solution:

Using formula,

$\bf{\dag}\:{\boxed{\underline{\underline{\bf{\blue{S_{n} = \dfrac{n}{2}\Big[2a + (n - 1)d\Big]}}}}}}$

Where,

• '$\sf S_{n}$' denotes sum of first nth terms
• 'n' denotes number of terms
• 'a' denotes first term
• 'd' denotes common difference

Given,

• $\sf S_{8}$ = 136
• n = 8

Also,

• $\sf S_{12}$ = 300
• n = 12

Putting all values in formula we get,

$\tt \longrightarrow\:S_{8} = {\cancel{\dfrac{8}{2}}}\Big[2a + (8 - 1)d\Big]$

$\tt \longrightarrow\:136 = 4\Big[2a + 7d\Big]$

$\tt \longrightarrow\:2a + 7d = {\cancel{\dfrac{136}{4}}}$

$\longrightarrow\:{\underline{\underline{\tt{\purple{2a + 7d = 34}}}}}\qquad\tt{- Eq^{n}\:(1)}$

Also,

$\tt \longrightarrow\:S_{12} = {\cancel{\dfrac{12}{2}}}\Big[2a + (12 - 1)d\Big]$

$\tt \longrightarrow\:300 = 6\Big[2a + 11d\Big]$

$\tt \longrightarrow\:2a + 11d = {\cancel{\dfrac{300}{6}}}$

$\longrightarrow\:{\underline{\underline{\tt{\purple{2a + 11d = 50}}}}}\qquad\tt{- Eq^{n}\:(2)}$

Using formula,

$\bf{\dag}\:{\boxed{\underline{\underline{\bf{\red{a_{n} = a + (n - 1)d}}}}}}$

Where,

• '$\sf a_{n}$' denotes nth term
• 'a' denotes first term
• 'n' denotes number of terms
• 'd' denotes common difference

❶

• We have to find Sum of first and 8th terms.

i.e,

$\tt \longrightarrow\:a + a_{8}$

$\tt \longrightarrow\:a + \Big[a + (8 - 1)d\Big]$

$\tt \longrightarrow\:a + a + 7d$

$\tt \longrightarrow\:2a + 7d$

As from Eqⁿ (1) we know that 2a + 7d = 36. Therefore,

$\longrightarrow\:{\underline{\boxed{\bf{a + a_{8} = 36}}}}$

❷

• We have to find Sum of first and 12th terms.

i.e,

$\tt \longrightarrow\:a + a_{12}$

$\tt \longrightarrow\:a + \Big[a + (12 - 1)d\Big]$

$\tt \longrightarrow\:a + a + 11d$

$\tt \longrightarrow\:2a + 11d$

As from Eqⁿ (2) we know that 2a + 11d = 50. Therefore,

$\longrightarrow\:{\underline{\boxed{\bf{a + a_{12} = 50}}}}$

❸

• We have to find the number got by adding three times the first term to the 19th term.

i.e,

$\tt \longrightarrow\:3a + a_{19}$

$\tt \longrightarrow\:3a + \Big[a + (19 - 1)d\Big]$

$\tt \longrightarrow\:3a + a + 18d$

$\tt \longrightarrow\:4a + 18d$

$\tt \longrightarrow\:2a + 2a + 7d + 11d$

$\tt \longrightarrow\:(2a + 7d) + (2a + 11d)$

As from Eqⁿ (1) and Eqⁿ (2) we know that 2a + 7d = 36 and 2a + 11d = 50. Therefore,

$\tt \longrightarrow\:Required\:number = 34 + 50$

$\longrightarrow\:{\underline{\boxed{\bf{Required\:number = 84}}}}$

Hence,

• ❶ Sum of first and 8th terms is 34.
• ❷ Sum of first and 12th term is 50.
• ❸ The number got by adding three times the first term to the 19th term is 84.

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