Question

Help please. been stuck on this for the past hour!!

Answer 1

• A) 100 terms, B) 63 terms, C) 101 terms

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Given sequences:

• A) 1, 4, 4², ..., 4⁹⁹
• B) 11, 15, 19, 23, ..., 259
• C) 44, 45, 46, ..., 144

To find

• The number of terms.

Solution

A) We know any number with power of zero equals 1, therefore the sequence can be expressed as powers of 4:

• 4⁰, 4¹, 4², 4³, ... , 4⁹⁹

The number of terms from zero to 99 is:

• 99 + 1 = 100

B) This is a AP with the first term of 11 and common difference of 4.

The nth term equation is:

• aₙ = a₁ + (n - 1)d

Substitute the values and solve for n:

• 259 = 11 + (n - 1)*4
• 4(n - 1) = 259 - 11
• 4(n - 1) = 248
• n - 1 = 248/4
• n - 1 = 62
• n = 63

C) Same as above, the sequence is AP, with the first term 44 and common difference 1.

Applying the nth term formula:

• 144 = 44 + (n - 1)*1
• n - 1 = 144 - 44
• n - 1 = 100
• n = 101