∵a1,a2,a3,…,an are in HP ∴a11,a21,a31,…,an1 are in AP
Let d be the common difference of AP. ∴a21−a11=d⇒a1−a2=a1a2d
Similarly a2−a3=a2a3d …………… …………… an−1−an=an−1and
On adding all of these, we get a1−an=d{a1a2+a2a3+……+an−1an}… (i)
Also an1=a11+(n−1)d ⇒d=a1an(n−1)a1−an
On putting this value of d in equation (i), we get a1−an=a1an(n−1)a1−an{a1a2+a2a3+……+an−1an} ⇒a1a2+a2a3+…+an−1an=a1an(n−1)