Question

solve this
chapter progressions​

Answer 1

14)

Given :-

• a = First Term = (17/2)
• d = common Difference = (3/2)
• n = No. of Terms of AP = 64 .

using AP sum formula we get,

→ Sn = (n/2)[2a + (n - 1)d]

Putting All values we get,

→ Sn = (64/2)[2*17/2 + (64-1)(3/2)]

→ Sn = 32[ 17 + (63*3)/2 ]

→ Sn = 32[ 17 + 94.5 ]

→ Sn = 32 * 111.5

→ Sn = 3568 (Option 3) (Ans.)

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15)

Given :-

• l = last Term = 8
• sum of 8 terms = (-20)
• n = No. of Terms of AP = 8

using AP sum formula we get,

→ Sn = (n/2)[First term + Last Term]

Putting All values we get,

→ (-20) = (8/2)[ a + 8 ]

→ (-20) = 4[ a + 8 ]

→ (-5) = a + 8

→ a = -5 - 8

→ a = (-13) (Option 2) (Ans.)

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16)

Given :-

• First Term = a = 24
• Common Diff. = a2 - a1 = 20 - 24 = (-4)
• Sn = 72 .

using AP sum formula we get,

→ Sn = (n/2)[2a + (n - 1)d]

Putting All values we get,

→ 72 = (n/2)[2*24 + (n-1)(-4)]

→ 72*2 = n[48 - 4n + 4 ]

→ 144 = n(52 - 4n)

→ 144 = 4n(13 - n)

→ 36 = n(13 - n)

→ 36 = 13n - n²

→ n² - 13n + 36 = 0

→ n² - 9n - 4n + 36 = 0

→ n(n - 9) - 4(n - 9) = 0

→ (n - 9)(n - 4) = 0

→ n = 9 & 4. (Option 4) (Ans.)

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