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) \(7 + 10\frac 12 + 14 + …………+ 84\)
For this A.P.,
\(a = 7\) and \(l = 84\)
\(d = a_2-a_1 = 10\frac 12 -7 = \frac {21}{2} - 7 = \frac 72\)
Let 84 be the nth term of this A.P.
\(l = a + (n − 1)d\)
\(84 = 7 + (n-1)\frac 72\)
\(77 = (n-1)\frac 72\)
\(22 = n − 1\)
\(n = 23\)
We know that,
\(S_n = \frac n2[a+l]\)

\(S_{23} = \frac {23}{2}[7+84]\)

\(S_{23} = \frac {23 \times 91}{2}\)

\(S_{23} =\frac {2093}{2}\)

\(S_{23} = 1046\frac 12\)

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(ii) \(34 + 32 + 30 + ……….. + 10\)
For this A.P., \(a = 34\), \(d = a_2 − a_1 = 32 − 34 = −2\) and \(l = 10\)
Let \(10 \) be the n^th term of this A.P. 
\(l = a + (n − 1) d\)
\(10 = 34 + (n − 1) (−2)\)
\(−24 = (n − 1) (−2)\)
\(12 = n − 1\)
\(n = 13\)
\(S_n = \frac n2[a+l]\)

\(S_n = \frac {13}{2}[34+10]\)

\(Sn = \frac {13 \times 44}{2}\)
\(S_n = 13 \times 22\)
\(S_n = 286\)

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(iii) \((−5) + (−8) + (−11) + ………… + (−230)\)
For this A.P.,
\(a = −5\), \(l = −230\) and \(d = a_2 − a_1 = (−8) − (−5) = − 8 + 5 = −3\)
Let \(−230\) be the n^th term of this A.P.
\(l = a + (n − 1)d\)
\(−230 = − 5 + (n − 1) (−3)\)
\(−225 = (n − 1) (−3)\)
\(n − 1 = 75\)
\(n = 76 \)
And, \(S_n =\frac n2[a+l]\)

\(S_n = \frac {76}{2}[(-5)+(-230)]\)
\(S_n = 38(-235)\)
\(S_n = - 8930\)