(a) चूँकि ${a_1},\;{a_2},\,{a_3},………..,{a_n}$ एक समान्तर श्रेणी है। अत:  

 ${a_2} – {a_1} = {a_4} – {a_3} = ……. = {a_{2n}} – {a_{2n – 1}} = d$

 $\therefore $ $a_1^2 – a_2^2 + a_3^2 – a_4^2 + $$……. + a_{2n – 1}^2 – a_{2n}^2$

 $ = ({a_1} – {a_2})({a_1} + {a_2}) + ({a_3} – {a_4})({a_3} + {a_4}) + ……..$

 $…… + ({a_{2n – 1}} – {a_{2n}})({a_{2n – 1}} + {a_{2n}})$

  $ =  – d({a_1} + {a_2} + ……. + {a_{2n}}) =  – d\left\{ {\frac{{2n}}{2}({a_1} + {a_{2n}})} \right\}$

एवं हम जानते हैं कि ${a_{2n}} = {a_1} + (2n – 1)d$

$ \Rightarrow $ $d = \frac{{{a_{2n}} – {a_1}}}{{2n – 1}}$

$ \Rightarrow $ $ – d = \frac{{{a_1} – {a_{2n}}}}{{2n – 1}}$.

  $\therefore $ योगफल= $\frac{{n({a_1} – {a_{2n}}).({a_1} + {a_{2n}})}}{{2n – 1}} = \frac{n}{{2n – 1}}(a_1^2 – a_{2n}^2)$है।