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A sequence of numbers $a_1, a_2, a_3, \ldots$ is defined by $$a_{n+1} = 5a_n - 3, \ n > 1$$ Given that $a_2 = 7$, (a) find the value of $a_1$ (b) Find the value of $$\sum_{r=1}^{4} a_r$$ - Edexcel - A-Level Maths Pure - Question 6 - 2014 - Paper 1

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A-sequence-of-numbers-$a_1,-a_2,-a_3,-\ldots$-is-defined-by----$$a_{n+1}-=-5a_n---3,-\-n->-1$$----Given-that-$a_2-=-7$,---(a)-find-the-value-of-$a_1$---(b)-Find-the-value-of-$$\sum_{r=1}^{4}-a_r$$-Edexcel-A-Level Maths Pure-Question 6-2014-Paper 1.png

A sequence of numbers $a_1, a_2, a_3, \ldots$ is defined by $$a_{n+1} = 5a_n - 3, \ n > 1$$ Given that $a_2 = 7$, (a) find the value of $a_1$ (b) Find the ... show full transcript

Worked Solution & Example Answer:A sequence of numbers $a_1, a_2, a_3, \ldots$ is defined by $$a_{n+1} = 5a_n - 3, \ n > 1$$ Given that $a_2 = 7$, (a) find the value of $a_1$ (b) Find the value of $$\sum_{r=1}^{4} a_r$$ - Edexcel - A-Level Maths Pure - Question 6 - 2014 - Paper 1

Step 1

find the value of $a_1$

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Answer

Given the formula for the sequence, we can substitute n=2n=2:

a3=5a23a_3 = 5a_2 - 3
Substituting a2=7a_2 = 7:

a3=5(7)3=353=32a_3 = 5(7) - 3 = 35 - 3 = 32

Next, to find a1a_1, we substitute n=1n=1 in the original sequence formula:

a2=5a13a_2 = 5a_1 - 3
Substituting a2=7a_2 = 7:

7=5a137 = 5a_1 - 3
Solving for a1a_1 gives:

5a1=7+3=105a_1 = 7 + 3 = 10
a1=105=2a_1 = \frac{10}{5} = 2

Step 2

Find the value of $\sum_{r=1}^{4} a_r$

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Answer

We first need to determine the first four terms of the sequence:

  • We have already found a1=2a_1 = 2 and a2=7a_2 = 7.
  • From earlier, we calculated a3=32a_3 = 32. Now, let's find a4a_4 by substituting n=3n=3:

a4=5a33=5(32)3=1603=157a_4 = 5a_3 - 3 = 5(32) - 3 = 160 - 3 = 157

Now we have:

  • a1=2a_1 = 2
  • a2=7a_2 = 7
  • a3=32a_3 = 32
  • a4=157a_4 = 157

We can now compute the sum:

r=14ar=a1+a2+a3+a4=2+7+32+157\sum_{r=1}^{4} a_r = a_1 + a_2 + a_3 + a_4 = 2 + 7 + 32 + 157
Calculating the sum step by step:

  • 2+7=92 + 7 = 9
  • 9+32=419 + 32 = 41
  • 41+157=19841 + 157 = 198

Thus, the value of the sum is 198.

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