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Find the first four terms of the arithmetic progression, where π π denotes the sum of its first π terms, π five plus π seven is 167, and π 10 is 235.
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For an arithmetic progression with first term π and common difference π, the sum of its first π terms is π over two multiplied by two π plus π minus one multiplied by π.
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For its first five terms, thatβs five over two multiplied by two π plus five minus one π.
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Five minus one is four.
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So the inside of these brackets simplifies to two π plus four π.
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We can also divide though by two.
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And we get five multiplied by one π plus two π.
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And then, if we multiply each part of this bracket by the number on the outside, we end up with five π plus 10π.
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Letβs repeat this process for the first seven terms.
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We get seven over two multiplied by two π plus seven minus one π.
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Seven minus one is six.
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And we can once again divide through by two.
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Finally, when we expand the brackets by multiplying each term inside the bracket by seven on the outside, we get seven π plus 21π.
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We know that the sum of the first five terms plus the sum of the first seven terms is equal to 167.
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So we can add these two expressions and equate it to 167.
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Finally, we should simplify by collecting like terms.
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That gives us 12π plus 31π is equal to 167.
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Next, letβs substitute π equals 10 to form an equation for the sum of the first 10 terms.
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This time, we get 10 over two multiplied by two π plus 10 minus one multiplied by π.
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10 minus one is nine and 10 divided by two is five.
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Multiplying the brackets out and we get 10π plus 45π.
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We also know though that the sum of the first 10 terms is 235.
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So we can make this expression equal to 235.
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Notice that we now have two equations with two unknowns.
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We can solve these simultaneously to calculate the value of π and π.
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First, we need to make the coefficient of either π or π the same in each equation.
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Since 12 and 10 have a common multiple of 60, we can multiply the whole of the first equation by five and the whole of the second equation by six.
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That gives us 60π plus 155π equals 835 and 60π plus 270π equals 1410.
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Since the coefficient of π is now the same, we can subtract the first equation from the second.
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In fact, letβs change the order of these equations so itβs easier to see whatβs going on.
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60π minus 60π is zero, 270π minus 155π is 115π, and 1410 minus 835 is 575.
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To solve this equation, weβll divide both sides by 115.
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And that gives us that π is equal to five.
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We can substitute this value back into either of our equations to calculate the value of π.
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If we choose the second, that looks like this: 10π plus 45 multiplied by five equals 235.
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45 multiplied by five is 225.
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So we can begin to solve this equation by subtracting 225 from both sides.
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That gives us 10π is equal to 10, which means that π is equal to one.
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Weβve calculated the first term in our arithmetic progression is one and the difference between each term is five.
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All we need to do now is calculate the value of the first four terms of this sequence.
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Now, we could use the formula for the πth term in the sequence π plus π minus one π.
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However, we know that the first term is one and the difference between each term is five.
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So we can repeatedly add five to get the subsequent terms.
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The first term is one, the second term is six, the third term is 11, and the fourth term is 16.