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1. On 21st day of the month will she have 40 sweets left given that on 1st November Joan had 200 sweets in a box, she eats no sweets the first day and she eats 8 sweets per day from the box thereafter. This can be obtained by the formula of arithmetic sequence formula.

   On what day of the month will she have 40 sweets left?

   The arithmetic sequence formula of nth term of the sequence a, a+d, a+2d… is,

   ⇒ aₙ = a + (n – 1)d,

   where a is the first term, d is the common difference and d = aₙ – aₙ₋₁

   Here in the question it is given that,

   • on 1st November Joan had 200 sweets in a box

   • She eats no sweets the first day

   • she eats 8 sweets per day from the box thereafter

   We have to find that on what day of the month will she have 40 sweets left.

   From the given data we can write that,

   ⇒ Day 1 = 200 sweets

   ⇒ Day 2 = 192 sweets (8 sweets are eaten)

   ⇒ Day 3 = 184 sweets (8 sweets are eaten)

   …

   ⇒ Day n = 40 sweets

   Thus we can form an arithmetic sequence

   ⇒ 200, 192, 184,…,40 is an arithmetic sequence.

   Here, first term = a = 200

   Common difference = d = 192 – 200 ⇒ d = -8

   By using the formula, nth term will be,

   ⇒ a + (n – 1)d = 40

   ⇒ 200 + (n -1)(-8) = 40

   ⇒ 200 – 8n + 8 = 40

   ⇒ 208 – 8n = 40

   ⇒ 8n = 208 – 40

   ⇒ 8n = 168

   ⇒ n = 168/8

   ⇒ n = 21

   Hence on 21st day of the month will she have 40 sweets left given that on 1st November Joan had 200 sweets in a box, she eats no sweets the first day and she eats 8 sweets per day from the box thereafter.

   Learn more about arithmetic sequence here:

   brainly.com/question/16954227

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