Let an be an arithmetic progression with a4 + a7 + a10 = 17 and a4 + a5 +···+ a13 + a14 = 77. If ak = 13, then k =?

Answer:The value of k is 18.Step-by-step explanation:Here, the AP is,[tex]a_1, a_2,........a_n[/tex]Let a be the first term and d is the common difference,So the arithmetic sequence would be,[tex]a, a+d, a+2d, ..........a+(n-1)d[/tex]Given,[tex]a_4 + a_7 + a_10 = 17[/tex][tex]\implies a+3d+a+6d+a+9d=17[/tex][tex]3a+18d=17------(1)[/tex]Now, [tex]a_4 + a_5 +.......+ a_{13}+ a_{14}= 77[/tex][tex]\implies \frac{11}{2}(2(a+3d)+(11-1)d)=77[/tex][tex]11(2a+6d+10d)=154[/tex][tex]22a+176d=154-----(2)[/tex]22 × equation (1) - 3 × equation (2),We get,[tex]396d-528d = 374 - 462[/tex][tex]-132d=-88[/tex][tex]\implies d=\frac{88}{132}=\frac{2}{3}[/tex]From equation (1),[tex]3a+\frac{36}{3}=17[/tex][tex]3a+12=17[/tex][tex]3a=5[/tex][tex]a=\frac{5}{3}[/tex]Here,[tex]a_k=13[/tex][tex]a+(k-1)d=13[/tex][tex](k-1)d=13-a[/tex][tex]k-1=\frac{13-a}{d}[/tex][tex]k=\frac{13-a}{d}+1[/tex]By substituting the value,[tex]k=\frac{13-\frac{5}{3}}{\frac{2}{3}}+1=\frac{39-5}{2}+1=17+1=18[/tex]Hence, the value of k is 18.