Question

Find the sums given below :

1. \(7 + 10\frac 12+ 14 + ....... + 84\)
2. \(34 + 32 + 30 + ....... + 10\)
3. \(–5 + (–8) + (–11) + ....... + (–230)\)

Answer 1

(i) Calculate the sum of the arithmetic progression: \(7 + 10\frac 12 + 14 + …………+ 84\).
For this A.P., the first term is \(a = 7\) and the last term is \(l = 84\).
The common difference is \(d = a_2-a_1 = 10\frac 12 -7 = \frac {21}{2} - 7 = \frac 72\).
Let 84 be the nth term of this A.P.
Using the formula \(l = a + (n − 1)d\):
\(84 = 7 + (n-1)\frac 72\)
\(77 = (n-1)\frac 72\)
\(22 = n − 1\)
\(n = 23\)
The sum of the first n terms of an A.P. is given by \(S_n = \frac n2[a+l]\).
\(S_{23} = \frac {23}{2}[7+84]\)
\(S_{23} = \frac {23 \times 91}{2}\)
\(S_{23} =\frac {2093}{2}\)
\(S_{23} = 1046\frac 12\)

(ii) Calculate the sum of the arithmetic progression: \(34 + 32 + 30 + ……….. + 10\).
For this A.P., the first term is \(a = 34\), the common difference is \(d = a_2 − a_1 = 32 − 34 = −2\), and the last term is \(l = 10\).
Let 10 be the nth term of this A.P.
Using the formula \(l = a + (n − 1) d\):
\(10 = 34 + (n − 1) (−2)\)
\(−24 = (n − 1) (−2)\)
\(12 = n − 1\)
\(n = 13\)
The sum of the first n terms of an A.P. is given by \(S_n = \frac n2[a+l]\).
\(S_n = \frac {13}{2}[34+10]\)
\(Sn = \frac {13 \times 44}{2}\)
\(S_n = 13 \times 22\)
\(S_n = 286\)

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(iii) Calculate the sum of the arithmetic progression: \((−5) + (−8) + (−11) + ………… + (−230)\).
For this A.P., the first term is \(a = −5\), the last term is \(l = −230\), and the common difference is \(d = a_2 − a_1 = (−8) − (−5) = − 8 + 5 = −3\).
Let −230 be the nth term of this A.P.
Using the formula \(l = a + (n − 1)d\):
\(−230 = − 5 + (n − 1) (−3)\)
\(−225 = (n − 1) (−3)\)
\(n − 1 = 75\)
\(n = 76\)
The sum of the first n terms of an A.P. is given by \(S_n =\frac n2[a+l]\).
\(S_n = \frac {76}{2}[(-5)+(-230)]\)
\(S_n = 38(-235)\)
\(S_n = - 8930\)