Question

50 points
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Answer 1

☯︎ ᴀɴsᴡᴇʀ

• n = term number
• Sn = sum of first n terms
• a = First term
• an = nth term
• l = last term
• d = common difference

here, we will use following fomulas,

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

$\underline{ \boxed{ \red{ \bold{S_{n} = \frac{n}{2} \{2a + (n - 1)d \}}}}}$

${\underline{ \boxed{ \red{ \bold{S_{n} = \frac{n}{2} (a + l) }}}}}$

_________________________

viii) to find : value of first term and n

we have,

• an = 4
• d = 2

$\bold{: \implies 4 = a + (n - 1)2 } \: \: \: \: \\ \\ \bold{ : \implies 4 = a + 2n - 2} \: \: \: \: \: \: \: \: \\ \\ \bold{ : \implies 4 + 2 = a + 2n} \: \: \: \: \: \: \: \: \\ \\ \bold{ : \implies a + 2n = 6} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bold{ : \implies a = 6 - 2n}........(i)$

and

we have,

• Sn = -14
• d = 2

$\bold{: \implies - 14 = \frac{n}{2} \{2a + (n - 1)2 \} } \: \: \: \: \: \: \: \\ \\ \bold{ : \implies - 14 = \frac{n}{ \cancel2} \cancel2(a + n - 1)} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bold{ : \implies - 14 = n(a + n - 1).........(ii)}$

we get two equations,

• a = 6 - 2n .........(i)
• -14 = n(a + n -1) ........(ii)

now, substitute the value of a from (i) in (ii)

➪ -14 = n ( 6 - 2n + n - 1 )

➪ -14 = n ( 5 - n )

➪ - 14 = 5n - n²

➪ n² - 5n - 14 = 0

now, solve this quadratic equation.

n² - 5n - 14 = 0

➪ n² - 7n + 2n - 14 = 0

➪ n(n - 7) + 2(n - 7) = 0

➪ (n - 7) (n + 2) = 0

➪ n - 7 = 0 and n + 2 = 0

➪ n = 7 and n = -2(not possible)

now, substitute the value of n in (i)

➪ a = 6 - 2 × 7

➪ a = 6 - 14

➪ a = -8

hence, value of n is 7 and a is -8

__________________________

ix) to find : common difference

we have,

• a = 3
• n = 8
• Sn = 192

$\bold{ : \implies 192 = \cancel \frac{8}{2} \{2 \times 3+ (n - 1)d \}} \\ \\ \bold{ : \implies192 = 4 \{3 + 3 + (n - 1)d \} } \\ \\ \bold{ : \implies \cancel\frac{192}{4} = 3 + a_{n} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bold{ : \implies 48 = 3 +a_{n} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bold{ : \implies a_{n} = 48 - 3} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bold{: \implies \boxed{ \bold{ a_{n} = 45}}} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

now,

an = a + (n - 1)d

➪ 45 = 3 + (8 - 1)d

➪ 45 = 3 + 7d

➪ 7d = 45 - 3

➪ 7d = 42

➪ d = 42/7

➪ d = 6

so, value of d is 6

__________________________

x) to find : first term

we have,

• l = 28
• Sn = 144
• n = 9

$\bold{: \implies 144 = \frac{9}{2}(a + 28) } \: \\ \\ \bold{ : \implies \frac{ \cancel{144} \times 2}{ \cancel{9}} = a + 28} \\ \\ \bold{ : \implies a + 28 = 16 \times 2 } \: \: \: \\ \\ \bold{ : \implies a = 32 - 28} \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bold{ : \implies \boxed{ \bold{a = 4}}} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

so, value of a is 4

Answer 2

☯︎ ᴀɴsᴡᴇʀ

• n = term number
• Sn = sum of first n terms
• a = First term
• an = nth term
• l = last term
• d = common difference

here, we will use following fomulas,

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

$\underline{ \boxed{ \red{ \bold{S_{n} = \frac{n}{2} \{2a + (n - 1)d \}}}}}$

${\underline{ \boxed{ \red{ \bold{S_{n} = \frac{n}{2} (a + l) }}}}}$

_________________________

viii) to find : value of first term and n

we have,

• an = 4
• d = 2

$\bold{: \implies 4 = a + (n - 1)2 } \: \: \: \: \\ \\ \bold{ : \implies 4 = a + 2n - 2} \: \: \: \: \: \: \: \: \\ \\ \bold{ : \implies 4 + 2 = a + 2n} \: \: \: \: \: \: \: \: \\ \\ \bold{ : \implies a + 2n = 6} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bold{ : \implies a = 6 - 2n}........(i)$

and

we have,

Sn = -14

d = 2

$\bold{: \implies - 14 = \frac{n}{2} \{2a + (n - 1)2 \} } \: \: \: \: \: \: \: \\ \\ \bold{ : \implies - 14 = \frac{n}{ \cancel2} \cancel2(a + n - 1)} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bold{ : \implies - 14 = n(a + n - 1).........(ii)}$

we get two equations,

a = 6 - 2n .........(i)

-14 = n(a + n -1) ........(ii)

now, substitute the value of a from (i) in (ii)

➪ -14 = n ( 6 - 2n + n - 1 )

➪ -14 = n ( 5 - n )

➪ - 14 = 5n - n²

➪ n² - 5n - 14 = 0

now, solve this quadratic equation.

n² - 5n - 14 = 0

➪ n² - 7n + 2n - 14 = 0

➪ n(n - 7) + 2(n - 7) = 0

➪ (n - 7) (n + 2) = 0

➪ n - 7 = 0 and n + 2 = 0

➪ n = 7 and n = -2(not possible)

now, substitute the value of n in (i)

➪ a = 6 - 2 × 7

➪ a = 6 - 14

➪ a = -8

hence, value of n is 7 and a is -8

__________________________

ix) to find : common difference

we have,

• a = 3
• n = 8
• Sn = 192

$\bold{ : \implies 192 = \cancel \frac{8}{2} \{2 \times 3+ (n - 1)d \}} \\ \\ \bold{ : \implies192 = 4 \{3 + 3 + (n - 1)d \} } \\ \\ \bold{ : \implies \cancel\frac{192}{4} = 3 + a_{n} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bold{ : \implies 48 = 3 +a_{n} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bold{ : \implies a_{n} = 48 - 3} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bold{: \implies \boxed{ \bold{ a_{n} = 45}}} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

now,

an = a + (n - 1)d

➪ 45 = 3 + (8 - 1)d

➪ 45 = 3 + 7d

➪ 7d = 45 - 3

➪ 7d = 42

➪ d = 42/7

➪ d = 6

so, value of d is 6

__________________________

x) to find : first term

we have,

• l = 28
• Sn = 144
• n = 9

$\bold{: \implies 144 = \frac{9}{2}(a + 28) } \: \\ \\ \bold{ : \implies \frac{ \cancel{144} \times 2}{ \cancel{9}} = a + 28} \\ \\ \bold{ : \implies a + 28 = 16 \times 2 } \: \: \: \\ \\ \bold{ : \implies a = 32 - 28} \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bold{ : \implies \boxed{ \bold{a = 4}}} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

so, value of a is 4